Jee

ATOMIC PHYSICS

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MASS OF A PHOTONRest mass of photon is zero.Effective mass : If we assume that m is the effective mass of photons in moving conditions, then accordingto Einsteins theory.Energy of photon, E = mc2

m = 2Ec

but E = h = pc, p = photon momentum

m = 2Ec = 2

hc

= pc

C1: Draw the graph of (a) momentum versus effective mass of photon (b) Energy versus effective mass,(c) wavelength versus momentum, (d) Effective mass versus wavelength for a photon.

Sol:

p

mO

E

mO

(a) p = mc straight line tan = c (b) E = mc2 straight line tan = c2

p

m

(c)de-broglie wavelength of photon = hp (d) Effective mass of photon

p = h Rectangular hyperbola m = 2Ec = 2

hc

= 2hcc

m = hc Rectangular hyperbola

Power of a light sourceSuppose P = Power of the light source, = frequency of emitted photons,Energy of single photon = hLet, n = numbers of photons emitted per sec by the source, thenThus, Energy emitted per sec. by the source = W = h

Intensity of light source at a pointThe amount of energy incident on a point per unit area at that point in a unit time is called the intensity of lightat that point.

ATOMIC PHYSICS


Thus, intensity at a point is energy incident per unit area unit time.Suppose, I = Intensity of light at a point

= Frequency of photonn0 = no. of photon incident / sec / Area

I = n0hSI unit of intensity is J/second m2.

Intensity of light at a distance r :Let, W = Power of a point source

rPAmount of energy received by spherical surface per second

= Amount of energy emitted by light source per second = WHence, Intensity at distance

r = W

Area = 2W

4 r

Thus, I = 2W

4 r

Example 1: The intensity of sunlight on the surface of earth is 1400 W/m2. Assuming the mean wavelength ofsunlight to be 6000 , calculate the number of photons emitted from the sun per second assuming theaverage radius of earths orbit around sun is 1.49 1011 m.

Sol: Average radius, r = 1.49 1011 m. intensity of sunlight received by earth = I = 1400 W/m2, = mean wavelength = 6000 .

Energy emitted per second by the sum = Power of the sun = Wbut, Power, W = nh, where n is number of photons emitted by sun per sec. .....(i)Again, intensity at a distance r from a point source of power W.

I = 2W

4 r

rSun Earth

Power = P

I = 2nh4 r

[From equation (i)]

n =2I 4 r

h

= 2I 4 r

hc

[ h = hc

]

n =24 r I

hc

= 2 22 7

34 8

4 (1.49) 10 6 10 14006.63 10 3 10

n = 1.18 1045 photons per sec.

Photon Flux (i.e. photon/sec)Suppose, W = Power of a source.If A is the area of the metal surface on which radiations are incident, then the power received by the plate is

W = IA = 2W

4 r A

If is the frequency of radiation, then the energy is photon is given by h.

ATOMIC PHYSICS


If n is the number of photons incident on the plate per second, then, we haveW = n h

2

W4 r

A = N h

r

PArea(A)

Plate n = 2

W A4 r

h

n is called photon flux.

Photon flow density (n0) :The number of photons incident on unit area of the plate in one second is called the photon flow density.

i.e., Photon flow density, n0, = n 'A = 2

W4 r h

Photon-concentration :The number of photons per unit volume of the space.If n is the number of photons incident per unit area per second, then

Photon - concentration = nc , where c is the speed of light.

Example 2: The power of light emitted by the sun is 3.9 1026 W. Assuming the mean wavelength of sunlight tobe 6000 , calculate the photon flux arriving at 1 m2 area on earth perpendicular to light radiations. Theaverage radius of earths orbit around sun is 1.49 1011 m.

Sol: Power of light emitted by the sun, W = 3.9 1026 Wr = earth sun mean distance = 1.49 1011 m = mean wavelength of sun light = 6000 = 6 107 m

rSun EarthPIntensity of light an earth, I = 2W

4 rPower received by area A on earth, W = I A

W = 2W

4 r AAbut, W = n h, where n is the number of photon incident per sec or photon-flux.

n = W'h =

2W A4 rh

n = 2W A

4 r hc

[h = hc/]

n = 26 7

2 22 34 8

(3.9 10 ) (6 10 ) 14 (1.49) 10 6.63 10 3 10

photon /sec

= 4.22 1021 photon/sec

ATOMIC PHYSICS


PRESSURE EXERTED BY PHOTONS OR RADIATIONS PRESSURE(a) Each photon carries energy and momentum. Hence when photons of light is incident on a surface, the light

is either absorbed or reflected or both is done by the surface.(b) change in momentum of light takes place, which causes impulse or force on the surface resulting in a

pressure called radiation pressure.

Let light of intensity I is incident on a surface. Each photon carries energy h and momentum = hc

.

Energy incident on unit area in unit time = I(by the definition of intensity).Number of photons incident on unit area in unit time.

i.e.N

A t = I

h , .....(i)

Momentum delivered to unit area in unit time,

i.e.,P

A t

= N

A t

(change in momentum of each photon) .....(ii)

Using equation (i) & (ii), we get

PA t

= I

h

(change in momentum of each photon)

but,P

A t

= Force exerted

A , [ force exerted = Pt

]

= Radiation pressure

Pr = Radiation pressure = I

h (change in momentum of each photon) .....(iii)

Radiation pressure for perfectly absorbing surface :

In this case, change of momentum suffered by each photon = hc

.

Using the equation (iii), we get

hvc Perfectly

absorbingsurface

Radiation pressure = I

h hc

= Ic

Radiation pressure = Ic for a perfectly absorbing surface.

Radiation pressure for a perfectly reflecting surface :

In this case, change in momentum of each photon = hc

hc

=

2hc

hvc Perfectly

reflectingsurface

hvcHence, using the equation (iii), we get

Radiation pressure = I

h

2h

c

= 2Ic

Radiation pressure = 2Ic for a perfectly reflecting surface.

ATOMIC PHYSICS


Radiation pressure for a surface of reflection coefficient () :

hvc surface

hvc

In this case, Momentum of incident photon = hc

Momentum of reflected photon = hc

, = reflection coefficient

Change in momentum = hc

hc

= (1 + )

hc

Thus, using equation (iii), we get,

Radiation pressure = (1 + )Ic for photons falling normally on a surface.

NOTE : (i) For a perfectly absorbing surface = 0

Radiation pressure = Ic

(ii) For a perfectly reflecting surface = 1

Radiation pressure = (1 + 1)Ic =

2Ic

C2: A laser emits a light pulse of duration = 0.13 ms and energy E = 10J. Find the mean pressure exerted bysuch a light pulse when it is focussed into a spot of diameter d = 10 m on a surface perpendicular to thebeam and possessing a reflection -coefficient = 0.50.

Sol: Laser energy, E = 10J

Laser d = 10 mdPlate area = 4

d2

Pulse duration, = 0.13 sPressure exerted by light pulse i.e.,

Radiation pressure = (1 + )Ic , by the definition of radiation pressure by photons of light .....(i)

Here, = reflection coefficient = 0.50I = Intensity of lightc = speed of light = 3 108 m/s

By the definition of intensity of light, I = Laser energy

(pulse duration) (Area of plate)

i.e. I = 2

E

d4

.....(ii)

From (i) & (ii) Radiation pressure = (1 + ) 24E

c d

= (1 + 0.50) 8 3 124 10

3 10 (0.13 10 ) 100 10

= 4.9 107 Nm2 = 490 atm

ATOMIC PHYSICS


Example 3: 1 A current flows in an x-ray tube operated at 10,000 V. The target area is 104 m2. Find thepressure on the target, assuming that the electrons strike the target normally and the photons leave the targetnormally. Consider the ideal situation where each incident electron gives rise to a photon of the same energy.

Sol: Energy of each electron = energy of each photon = E = 104 eVMomentum delivered by each electron = p1 = 2mE

(photon)

p1e

p2

Plate

Momentum taken away by each photon = p2 = Ec

Change of momentum due to each electron-photon pair = p1 (p2) = p1 + p2Current incident on target = i = 106A Number of electrons incident per second = i/e = no. of photons emitted per second.

Total momentum change per second = force = ie

(p1 + p2)

Pr, Pressure = force / area =i

Ae

(p1 + p2)

=i

Ae

E2mEc

Pr = 6

4 19

1010 1.6 10

4 1931 4 19 1/ 2

8

10 1.6 10(2 9.1 10 10 1.6 10 )3 10

= 3.7 106 Nm2

FORCE EXERTED BY PHOTONS ON A SURFACEWe know that, when photons of light is incident on a surface, the change in momentum of photons takesplace resulting in a force exertion by the photons on the surface.Let light of power W is incident on a surface.

Each photon carries energy h and, momentum = hc

.....(i)

Power = W SurfaceEnergy incident in unit time = W(by the definition of power)But, power is given by

W =N

t (h), where N = total no. of photons incident in time t.

Nt =

Wh .....(ii)

Momentum delivered in unit time.

Pt

=N

t

(change in momentum of each photon)

=Ph (change in momentum of each photon)

but,Pt

= rate of change of momentum

ATOMIC PHYSICS


Hence,Pt

= F = Force exerted by the photons.

Thus, force = Ph

(change in momentum of each photon) ...(iii)

Force exerted by photons on a perfectly absorbing surface :

In this case, change of momentum suffered by each photon = hc

using equation (iii), we gethvc Perfectly

absorbingsurface

force = Wh

hc

= Wc

thus, F = Wc for a perfectly absorbing surface.

Force exerted by photons on a perfectly reflecting surface:In this case, change of momentum suffered by

each photon = hc

hc

=

2hc hvc surface

hvc

using equation ( iii), we get

Force = Wh

2hc

= 2Wc

thus, F = 2Pc , for a perfectly reflecting surface.

Force exerted by photons on a surface of reflection coefficient :

In this case momentum of incident photon = hc

, hvc surface

hvc

Momentum of reflected photon = hc

Change in momentum = hc

hc

= (1 + )

hc

Thus, using equation (iii), we get

Force = h

(1 + )

hc

= (1 + )Wc

i.e. F = (1 + )Wc for photons falling normally on a surface.

ATOMIC PHYSICS


C3: If a point source of light of power W is placed at the centre of curvature ofa hemispherical surface, whose inner surface is completely reflecting, then

the force on the hemisphere due to the light falling on it is given by F = W2c .

W(Source)

C4: If a perfectly reflecting solid sphere of radius r is kept in the path of aparallel beam of light of large aperture and having an intensity I, then the

force exerted by the beam on the sphere is given F = 2r I

c.

I

r

Note that force is equal to the product of (I/c) and the projected area ofthe sphere.

C5: If photon of light of power W falls at an angle to a perfectly reflecting

surface, then net force exerted on the plate is given by Fnet = 2Wc cos.

C6: A laser emits a light pluse of duration T = 0.10 ms and energy E = 10 J. Find the mean pressure exerted bysuch a pulse when it is focused on a spot of diameter d = 10 m on a surface perpendicular to the beam andwith a reflection coefficient = 0.5.

Sol. p, momentum of a photon = hv/c = E/cMomentum of reflected photon = (E/c) change of momentum = (E/c) (E/c) = (1 + )E/cForce exerted = [(1 + ) E/c]/TPressure = 4 [(1 + ) E/c]/Td2

= 2 8 4 104(1 )E 4(1 0.5) 10

cTd 3 10 10 10

= 6.37 106 Nm2

C7: A plane light wave of intensity I = 0.50 W cm2 falls on a plane mirror of reflection coefficient = 0.8. Theangle of incidence = 45. Find the normal presure exerted by light on that surface.

Sol. If S is the area of the mirror on which light falls, the transverse section of the incident beam is S cos .Momentum of the incident photons = (I/c)(S cos ). Normal component of momentumflux = (IS cos /c) cos = IS cos 2/c.Momentum of reflected photons = (IS cos2/c) in the opposite direction. rate of change of momentum = force = (1 + ) IS cos2/c normal pressure = force/area = (1 + ) I cos2/c required pressure

= (1 + 0.5) (0.5 104 cos2 45)/3 108

= 1.25 105 Nm2

Example 4: An isotropic point source of radiation power P is placed on the axis of an ideal mirror. The distancebetween the source and the mirror is n times the radius of the mirro. Find the force that light exerts on themirror.

ATOMIC PHYSICS


Sol. Energy flux through the annular space between two cones of half-angles and + d

= (P/4) (2 sin d) momentum of flux = P sin d/2c

d

nRS

B

B

R

A

rate of change of momentum along the normal.2(P sin d/2c) cos = P sin cos d/c

force on mirror = 0

p sin cos dc

where = half-angle of the cone subtened by the mirror = tan1(1/n) F = P sin2 / 2c = P/2(n2 + 1)c

Example 5: Figure shows a small, plane strip of mass m suspended from a fixed support through a string. Acontinuous beam of monochromatic light is incident horizontally on the strip and is completely absorbed.The energy falling on the strip per unit time is W. find the deflection of the string from the vertical, if the stripstays in equilibrium.

Sol: Force exerted by the photons of light= (number of incident photons per sec) (change in momentum of each photon)

Let, Number of incident photons per sec = N

Light

change in momentum of each photon = hc

hence force exerted by the photons of light,

F = Nhc

= (Nh )

c

but, W = Nhv, by the definition of power

F =Wc

If the string makes an angle with the vertical when the strip is at equilibrium, then for the equilibrium of thestrip,

Tsin = F, in horizontal direction,

TF m

mg

T = tensionin the string

and Tcos = mg, in vertical direction.Dividing the above equations, we get

tan = F

mg = W/cmg =

Wcmg ,

tan = W

cmg = tan1

Wcmg

Example 6. A plane light wave of intensity I = 0.80 W cm2 illuminates a sphere of radius R = 5.0 cm. Find theforce that the light exerts on the sphere.

Sol. p, momentum of the incident pulse = E/c.

i p (E / c) (sin i cos j)

with respect to a frame of reference with the outward normal as the y-axis

and a line perpendicular to it and lying in the plane of the mirror as the x-axis.

ATOMIC PHYSICS


pf = E/c and f p ( E / c) (sin i cos j)

f i p p p ( 1)E / csin i ( 1) cos E / c j

2 2 2 2| p | E / c ( 1) sin ( 1) cos

2E / c 1 2 cos 2

8 2| p | 10 / 3 10 1 0.8 2 0.8cos60

= 5.2 108 Nm2

Impulse applied by photon on a surfaceLet hn be the energy of photons of a light incident normally on a surface,

Momentum of an incident photon = hc

.

Change in momentum of the photon takes place due to impact of the photons with the surface. This changein momentum of the photons causes impulse.Using impulse momentum theorem, we get

Impulse = Total change in momentum of photons = (Total number of photons) (change in momentum of each photon)

Impulse = N (change in momentum of each photon) ......(i)

Impulse on a totally absorbing surface:


using equation (i), we get

Impulse = N 'h

c

= Ec , where E is the total energy of the light

Impulse = Ec for a perfectly absorbing surface.

Impulse on a totally reflecting surface:Impulse = N (change in momentum of each photon)

= N h hc c

= 2(N 'h )

c

= 2Ec [ E = total energy = Nh]

Impulse = 2Ec for a perfectly reflecting surface.

Impulse on a surface of reflection-coefficient :


hc

= (1 + )

hc

.

Using equation (i), we get

ATOMIC PHYSICS


Impulse = N(1 + )hc

= (1 + )N 'h

c

= (1 + )Ec , [E = total energy]

Impulse = (1 + )Ec for photons falling normally on the surface.

Example 7: A small perfectly reflecting mirror of mass m = 10 mg issuspended by a weightless thread of length = 10 cm as shown in thefigure. Find the angle through which the thread will be deflected whena short laser with energy E = 13J is shot in horizontal direction at rightangles to the mirror. (g = 10 m/s2).

m

m

Laser

Sol: Impulse applies by laser on the mirror, Impulse = 2Ec , as mirror is perfectly

reflecting. Initial momentum of the mirror = 0

v = 0

v

(1 cos ) = H

Let, final momentum of the mirror = mvchange in momentum of the mirror due to impact = 0 (mv) = mv, wherev is the speed of mirror just after impact.But, Impulse momentum theorem gives,

Impulse = change in momentum2Ec = mv or v =

2Emc .....(i)

Total mechanical energy of the mirror will be conserved after impact.Thus, Loss in kinetic energy = Gain in potential energy

12 mv

2 = mgh v2 = 2gH

2

2 2

4Em c

= 2gH, [as v =2Emc from equation (i)]

2

2 2

2Em c

= gH 2

2 2

2Em c

= g(1 cos), [ H = (1 cos)]

1 cos = 2

2 2

2Em c g

2sin2 2

= 2

2 2

2Em c g [ 1 cos = 2sin

22

]

sin 2

= 2E

mc g Here E = 13J, m = 10 106 kg, = 0.1 m

sin 2

= 5 813

10 3 10 10 0.1 = 0.0043 or = 0.5

PHOTONS UNDER GRAVITY

ATOMIC PHYSICS


Photons can be considered as a particle of mass m.If m = mass of photon

v = frequency of photonE = energy of photon, then

mc2 = E = hv m = 2Ec = 2

hc

Thus, a photon of frequency v acts gravitationally like a particle of mass 2hc

.

C8: If a photon frequency v falls on the surface of earth from a height h, then what will be its frequency on thesurface of earth.

Sol: Change in frequency of the photon takes place. v, E = h + mgHv

v v, E = hGround

HLet v be the frequency of photon on the surface of earth.

Mass of photon = 2hc

Mass of the photon depends on its frequency, but we will consider the mass to be constant as difference inv and v is very small.Energy conservation gives

Initial energy = final energy hv + mgH = hv

hv + 2hc

gH = hv

v = v 2gH1c

Example 8: A planet of mass M and radius R emits a photon of frequency v. What will be the frequency ofphoton when it covers an infinite distance.

Sol: Let vbe the frequency photon at a infinite distance.vat R

M

vG = Gravitational constant

Mass of the photon, m = 2hc

Energy of photon on the surface = hv GMmR = hv 2

GM hR c

Energy of photon at infinity = hvEnergy must be conserved,

hv 2GM hR c

= hv v = v 2

GM1Rc

NOTE:

= 2GMRc is called frequency shift.

PHOTOELECTRIC EFFECT

ATOMIC PHYSICS


Ejection of electrons from a metal plate when illuminated by light or any other radiation of suitable wavelength(or frequency) is called photoelectric effect.This phenomenon was first discovered by Heinrich Hertz in 1887. One year later, Hallwachs made theimportant observation that when a negatively-charged zinc plate is irradiated with ultra-violet light, it losesits negative charge. Afterwards, it was discovered that alkali metals like lithium, sodium, potassium, rubidiumand caesium eject electrons when visible light falls on them. Ten years later, J.J. Thomson and P.Lenardshowed that the action of light was to cause the emission of free electrons from metal surface. Althoughthese electrons are no different from all other electrons, it is customary to refer to them as photoelectrons.

Experimental Study of Photoelectric EffectQuartz

C

A

+E

V

Photoelectric phenomenon can be studied with the help of a simpleapparatus shown in figure. It consists of two photosensitive surfacesE and C enclosed in an evacuated quartz bulb. In the absence oflight, there is no flow of current in the circuit and the ammeter Areads zero. When plate E is exposed to some source ofmonochromatic light, current starts flowing. However, no currentis found to flow when light falls on plate C.The explanation of the above behaviour lies in the fact that when E is irradiated with light, the incidentphotons eject electrons by collision with its atoms. These photoelectrons are immediately attracted by thecollector plate C thereby starting current flow as indicated by the ammeter.When C is irradiated, even then photoelectrons are produced but they are not allowed to leave plate C(i) firstly, because of the pulling effect of positive potential of C and(ii) secondly, due to repulsion from the negative plate E.Hence, no current is found to flow in the circuit.

Einsteins Photoelectric Equation:Following Plancks idea that light consists of photons, Einstein proposed an explanation of photoelectriceffect as early as 1905. According to this explanation when a single photon is incident on a metal surface, itis completely absorbed an imparts its energy hf to a single electron. The photon energy is utilised for twopurposes:(i) Partly for getting the electron free from the atom and away from the metal surface. This energy is knownas the photoelectric work function of the metal and is represented by W0.(ii) the balance of the photon energy is used up in giving the electron a kinetic energy of 1/2mv2.

hf = W0 + 1/2mv2 .....(i)

It is known as Einsteins photoelectric equation.In case, the photon energy is just sufficient to liberate the electron only then no energy would be availablefor imparting kinetic energy to the electron. Hence, the above equation would reduce to

hf0 = W0 .....(ii)where f0 is called the threshold frequency. It is defined as the minimum frequency which can cause photoelectricemission. For frequencies lower than f0, there would be no emission of electrons whereas for frequenciesgreater than f0, electrons would be ejected with a certain definite velocity (and hence kinetic energy).Substituting this value of W0 in equation (i) above, the Einsteins photoelectric equation becomes

hf = hf0 + 12 mv

2

or hf = hf0 + K.E.

Long Wavelength Limit (0)

ATOMIC PHYSICS


It is the wavelength corresponding to the threshold frequency f0. Its physical significance is that radiationshaving wavelength longer than 0 would not be able to eject electrons from a given material whereas thosehaving < 0 will. In other words, it represents the upper limit of wavelength for photoelectric phenomenon.By analogy, it is also referred to as threshold wavelength.

Now, c = f0 0 0 = 0

cf

Also, W0 = hf0 0

1f

= 0

hW

0 = 0

chW

(i) When W0 is in joules

0 = 8 34

0

3 10 6.625 10W

= 26

0

19.875 10W

metre

(ii) When W0 is electron volts (eV)

0 = 8 34

190

3 10 6.625 101.602 10 W

= 7

0

12.4 10W

metre =

0

12, 400W

Kinetic energy of PhotoelectronsEinsteins photoelectric equation can be used to find the velocity and hence the kinetic energy of an ejectedphoto-electron.

hf = W0 + 12 mv

2

= hf0 + K.E. K.E. = hf hf0 = h(f f0)

Now, f =c

and f0 = 0

c

K.E. = ch 0

1 1

= 3 108 6.625 10340

1 1

= 19.875 10260

1 1

joules and 0 in metres

= 19.875 1026 0

1 1

joules and 0 in

ATOMIC PHYSICS


=16

19

19.875 101.602 10

0

1 1

eV and 0 in

K.E. = 12,400 0

1 1

eV and 0 in

Incidentally, it may be noted that this also represents the maximum value of the kinetic energy a photoelectroncan have.

Emass = h(f f0) = h.f joules

= 12,400 0

1 1

eV and 0 in

Photoelectric Work FunctionAs explained above it is defined as the energy which is just sufficient to liberate electrons from a body withzero velocity. Its value is given by

W0 = hf0 = 0

ch =

8 34

0

3 10 6.625 10

=26

0

19.875 10 joules 0 in metres

=16

0

19.875 10 joules 0 in

=16

190

19.875 101.602 10

eV 0 in

W0 =0

12, 400 eV 0 in

Laws of Photoelectric Emission

RB

V

Quartz

CA

The apparatus shown in figure was used by Millikan to study the photoelectric effect and the various lawsgoverning it. S is a source of radiations of variable but known frequency f and intensity I. E is the emittingelectrode made of the photosensitive material and C is the collecting electrode. Both these electrodes areenclosed in an evacuated glass envelope with a quartz window that permits the passage of ultraviolet and

ATOMIC PHYSICS


visible light. As shown, any potential difference can be established between the two electrodes. A reversingswitch RS enables the polarities of the two electrodes to be reversed. If the tube is in the dark, then nophotoelectrons are emitted and the microammeter A read zero. However, if ultraviolet or visible light isallowed to fall on the emitting electrode, electrons are liberated and circuit current is set up.From the experimental data collected by Richardson and Compton in 1912, photoelectric emission wasfound to be governed by the following laws:(i) Photoelectric current (i.e., number of electrons emitted per second) is directly proportional to theintensity of the incident light (or radiation).

Light Intensity I

Phot

oele

ctric

curre

nt

Frequencyconstant

This can be verified by increasing the intensity of light andmeasuring the corresponding photoelectric current while holdingthe frequency of the incident light frequency of the incident lightand the accelerating potential V of the collecting electrode Cconstant. The graph is similar to one shown in figure. Increase inintensity means more photons and hence greater ejection ofelectrons.(ii) For each photosensitive surface, there is a minimum frequency ofradiation (called threshold frequency) at which emission begins.

f0Frequency

Phot

oele

ctric

curr

ent

Intensityconstant

Light of lower frequency (or longer wavelength), however strong,will not be able to produce any electron emission. This fact can beverified by keeping the light intensity constant while varying thefrequency. The graph so obtained is shown in figure. The current isfound to increase (though non-linearly) with the frequency of theincident light.Moreover, it is seen that there is a limiting or critical frequency belowwhich no photoelectons are emitted. It is called threshold frequencyand its value depends on the nature of the material irradiated because for each material there is a certainminimum energy nessecary to liberate an electron. This energy is known as photoelectric work function orthreshold energy W0. As seen from f0 = W0/h.The wavelength corresponding to be threshold frequency f0 is called threshold wavelength or long wavelengthlimit. No photo-electrons are emitted for wavelength greater than 0, no matter how long the light falls onthe surface or how greater is its intensity. The photoelectric or quantum yield (which is defined as the photo-electric current in amperes per watt of incident light) depends on the frequency (and not intensity) of theincident light.(iii) The maximum velocity of electron emission (and hence kinetic energy) varieslinearly with the frequency o the incident light but is independent of its intensity.As seen from Einsteins photoelectric equation of

2max

1 mv2 = h(f f0) or Emax f f0

Frequency

E max

Intensityconstant

Hence, increase in the frequency of the incident light increase the velocity with which photoelectrons areejected. The same fact is illustrated by figure. Incidentally, it may be noted that the slope of the curve givesthe value of Plancks constant h.If the intensity of the incident light is increased (keeping frequency constant), more photons will be incident

ATOMIC PHYSICS


on the metal surface, each photon having the same energy. Hence, more photo electrons will be ejected.Since an electron can absorb only one photon, each photoelectron will have the same energy and will beejected with the same velocity. Obviously, increase in intensity only increases the number of photo-electronsejected but not their kinetic energy.(iv) The velocities of emitted electrons have values between zero and a definite maximum. The proportionof the electrons having a particular velocity is independent of the intensity of radiation.(v) Electrons are emitted almost instantaneously even when the intensity of incident radiations is very low.The time lag between the incidence of radiation and emission of first electrons is less than 108 second.(vi) For a given metal surface, stopping potential V0 is directly proportional to frequency but is independentof the intensity of the incident light.

v

i

v0 O

a b I

HSuppose in figure, the frequency and intensity of incident light areheld constant but the potential difference V between the twoelectrodes E and C is increased. Up to some stage as this p.d. isincreased photoelectric current is also increased. However, soonsome value of V is reached when the current reaches a limiting orsaturation value when al the photoelectrons emitted by E areimmediately collected by C. Further increase in V hardly producesany appreciable increase in current as shown by the flat portion ofcurve I in figure.If V is reversed with the help of the reversing switch RS i.e. E is made positive and C negative, the currenti does not immediately drop to zero proving that electrons are emitted from E with some definite velocity.This velocity is such that it gives enough kinetic energy to the electrons so as to surmount the retardingelectric field between the two electrodes. Hence, some electrons do succeed in reaching C despite the factthat the electric field opposes their motion.When reversed V is made large enough, a value V0 (called stopping or inhibiting potential) is reached whencurrent is reduced to zero.Stopping potential is that value of the retarding potential difference between the two electrodes which is justsufficient to half the most energetic photoelectrons emitted.As seen from curve II of figure, doubling the intensity of the incident light merely doubles the current butdoes not affect the value of V0.Now, if vmax is the maximum velocity of emission of a photoelectron and V0 the stopping potential, then

2max

1 mv2 = eV0 or vmax =

02.eVm

or vmax =11

02 1.76 10 V

= 5.93 105 0V m/s Obviously Emax = eV0 joules

= V0 electron-voltIf however, the experiment is repeated by varying the frequency of the light, it is found that the stoppingpotential varies linearly with frequency as shown in figure. Below threshold frequency, no electrons areemitted, hence stopping potential is zero for that reason. But as frequency is increased above f0, the stoppingpotential varies linearly with the frequency of incident light.

ATOMIC PHYSICS


Einsteins photoelectric equation may be expressed in terms of stopping potential as given below:

hf = W0 + 2max

1 mv2

f0Frequency

Stop

ping

pot

entia

l, V

0

Now, W0 = hf0 and 2max

1 mv2 = eV0

hf = hf0 + eV0 or V0 = 0h(f f )

e

Now, f =c

and f0 = 0

c

V0 =che 0

1 1

= 12.4 107 0

1 1

volt and 0 in metres

V0 = 12,400 0

1 1

volt and 0 in

C9: Photoelectrons with a maximum speed of 7 105 m/sec are emitted from a metal surface when light offrequency 8 1014 Hz falls on it. Calyculate the threshold frequency of the surface.

Sol: Emax = h(f f0)

2max

1 mv2 = h(f f0)

or12 9.1 10

31 (7 105)2 = 6.625 1034 (8 1014 f0)

f0 = 4.635 1014 Hz

C10: A tungsten cathode whose threshold wavelength 2300 is irradiated by ultraviolet light of wavelength1800 . Calculate (i) the maximum energy of the photoelectrons emitted and (ii) the work function fortungsten, both in electron-volts.

Sol: (i) W0 =0

12, 400 =

12, 4002300 = 5.4 eV

(ii) Emax = 12,400 0

1 1

eV

= 12,400 1 1

1800 2300

= 1.5 eV

C11: If light of = 6000 falls on a metal surface and emits photoelectrons with a velocity of 4 105 m/s, whatis photoelectric threshold wavelength ?

ATOMIC PHYSICS


Sol: K.E. of photoelectrons = 12 9.1 10

31 5 2

19

(4 10 )1.6 10

= 0.445 eV

Energy content of photon of = 6000 = 12, 0006000 = 2.07 eV

W0 = 2.07 0.445 = 1.625 eV

0 = 12, 0001.625 = 7631

C12: Calculate the threshold frequency for gold having photoelectric work function equal to 4.8 eV. If light ofwavelength 2220 falls on gold, what will be maximum kinetic energy of the photoelectrons coming out?

Sol: Energy of the light photon = 12, 0002220 = 5.58 eV. Out of this, 4.8 eV would be used for dislodging the

electron and the balance would represent its kinetic energy.Emax = 5.58 4.8 = 0.78 eV

Alternatively, 0 = 12, 400

4.8 = 2583 . Hence, we may use

Emax = 12,400 0

1 1

eV..

C13: When violet light of = 4000 Au strikes the cathode of a photocell a retarding potential of 0.4 V isrequired to stop emission of electrons. Calculate (i) light frequency (ii) photon energy (iii) work function (iv)threshold frequency and (v) net energy after the electron leaves the surface.

Sol: (i) f = c

= 8

10

3 104000 10

= 7.5 1014 Hz(ii) E = hf = 6.625 1034 7.5 1014

= 4.95 1019 J = 3.1 eV(iii) W0 = hf K.E. = hf V0 = 3.1 0.4 = 2.7 eV

(iv) f0 = 0W

h = 19

34

2.7 1.6 106.625 10

= 6.5 1014 Hz(v) Net energy hf W0 = 3.1 2.7 = 0.4 eV = 6.4 10

20 J

Example 9: A photon of wavelength 3310 falling on a photo cathode ejects an electron of energy 3 1019J and one of wavelength 5000 ejects an electron of energy 0.972 1019 J. Calculate the value ofPlancks constant and the threshold wavelength for the photo cathode.

Sol: hf = W0 + K.E. or hc

= W0 + K.E.

ATOMIC PHYSICS


In the first case, 8

10

h 3 103310 10

= W0 + 3 1019

In the second case, 8

10

h 3 105000 10

= W0 + 0.972 1019

Subtracting one from the other,h = 6.62 1034 J-s

Substituting this value of h, W0 = 3 1019 J

0 = 0

chW =

8 34

19

3 10 6.62 103 10

= 6620 1010 m. vmax = 7.12 10

5 m/s

Example 10: A certain metal has a threshold wavelength of 6525 . Find the stopping potential when the metalis irradiated with(a) monochromatic light having a wavelength of 4000 .(b) light having twice the frequency and three times the intensity of that in (a) above.(c) If a material having double the work function were used, what would be the answer to (a) and (b)above?

Sol: (a) V0 = 12,400 1 1

4000 6525

= 1.2 volt

(b) Stopping potential is independent of the intensity of the incident light but varies directly as frequency fprovided it is more than f0.

Since frequency is twice, the wavelength of the light is half i.e. 2000

V0 = 12,400 1 1

2000 6525

= 4.3 V

(c) If work function is double, then 0 is reduced to half i.e. 0 = 6525

2 = 3262.5 . Since the incident light

has = 4000 , it would not be able to produce photoemission. In the second case, = 2000 .

V0 = 12,400 1 1

2000 3262.5

= 2.4 volt

C14: A certain metallic surface is illuminated by monochromatic light of variable wavelength. No photoelectronsare emitted above a wavelength of 5000 . With an unknown wavelength, a stopping potential of 3.1 V isnecessary to stop photoelectric current. Find the unknown wavelength.

Sol: Here, 0 = 5000 , V0 = 3.1 V, = ?

Now, V0 = 12,400 0

1 1

3.1 = 12,400 1 1

5000

= 2,222

ATOMIC PHYSICS


C15: Light of wavelength 2000 falls on an aluminium surface. In aluminium, 4.2 eV are required to remove anelectron. Determine (i) KE of the fastest emitted photo-electron (ii) KE of the slowest emitted photo-electron (iii) stopping potential and (iv) cut-off wavelength for aluminium.

Sol: Photon energy of the incident light is

= 124002000 = 6.2 eV

(i) Emax = (6.2 4.2) eV= 2 eV = 2 1.6 1019 = 3.2 1019 J

(ii) Emin = 0(iii) Ve = Emax = 2 V

(iv) 0 =0

12400W =

124004.2 = 2952.4

C16: The stopping potential is 4.6 V for light of frequency 2 1015 Hz. When light of frequency 4 1015 Hz isused, the stopping potential is 12.9 V. Calculate the value of Plancks constant.

Sol: eV0 = h(f f0)Substituting the two given values, we get

4.6e = h (2 1015 f0)12.9e = h(4 1015 f0)

Subtracting one from the other, we have8.3e = 2h 1015

8.3 1.6 1019 = 2h 1015

h =19

15

8.3 1.6 102 10

= 6.44 1034 Js

Example 11. 103 W of 5000 light is directed on a photoelectric cell. If the current in the cell is 0.16 A, thepercentage of incident photons which produce photoelectrons, is(A) 0.4% (B) .04% (C) 20% (D) 10%

Sol. The percentage of incident photons which produce photoelectrons is

= n / t 100N / t

...(1)

neI

t

16

19n I 0.16 10t e 1.6 10

12n 10

t

...(2)

andN hcW

t

16N W 10

t hc 4

...(3)

Persentage = 12

1610

4 10010

[from eq. (1), (2) and (3)]

Persentage 0.04% Hence (B) is correct.

ATOMIC PHYSICS


Example 12. In a photo-emissive cell, with exciting wavelength , the maximum kinetic energy of electron is K.

If the exciting wavelength is changed to 34

the the kinetic energy of the fastest emitted electron will be :

(A) 3K/4 (B) 4K/3 (C) less than 4K/3 (D) greater than 4K/3

Sol.hc K

...(1)

hc K3 / 4

...(2)

Substracte eqn. (1) from eq. (2)4hc hc K K3

4hc 3hc K K3

hc K K3

K K K3

KK K3

4KK

3 Hence (D) is correct.

Example 13. Let K1 be the maximum kinetic energy of photoelectrons emitted by a light of wavelength 1 andK2 corresponding to 2. If 1= 22, then :

(A) 2K1 = K2 (B) K1 = 2K2 (C) K1 < 2K

2 (D) K1 > 2K2

Sol. K.E. = 2P

2m

P 2mK

P K

h K

1K

1 2

2 1

KK

ATOMIC PHYSICS


1 = 22

2 2

2 1

2 KK

2

1

K4

K

21

KK4

21KK2

Hence (C) is correct.

Example 14. Radiation of two photon energies twice and five times the work function of metal are incidentsucessively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted is the twocases will be(A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1

Sol. E1 = 2 E2 = 5 E1 = + K12 = + K1K1 =

and E2 = + K25 = + K2K2 = 4

1

2

K 1K 4

2max 1

2max 2

1 mv 121 4mv2

2max12max 2

v 1v 4

max 1

max 2

v 1v 2

Hence (A) is correct.

Example 15. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons havemaximum kinetic energy Ta eV and de-Broglie wavelength a. the maximum kinetic energy of photoelectronsliberated from another metal B by photones of energy 4.7 eV is Tb = (Ta 1.5) eV. If the De-Brogliewavelength of these photoelectrons is b = 2a, then find(a) The work function of a (b) The work function of b is (c) Ta and Tb

ATOMIC PHYSICS


Sol. 2A Amax1h mv2

and aA

hP

Aa

hP

2

2 AA max

P1mv

2 2m where m is mass of electron.

2A

AP

h2m

or 4.25 = 2

A 2a

h2m

...(1)

For B,4.7 = B + Tb

2

B 2b

h4.7

2m

...(2)

But b = 2aand Tb = Ta 1.5After solving(a) 2.25 eV (b) 4.2 eV (c) 2 eV and 0.5 eV

C17: An isolated metal body is illuminated with monochromatic light and is observed to become charged to asteady positive potential 1.0 V with respect to the surrounding. The work function of the metal is 3.0 eV. thefrequency of the incident light is __________.

Sol. h = + eVor h = 3 + 1 = 4 eV

19

344 eV 4 1.6 10

h 6.63 10

= 0.96 1015 Hz

Example 16. 663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5 109 incident photons is absorbed and causes an electron to be ejected from the surface, the total photocurrentin the circuit is _______.

Sol.N

t = no. of photon incident per second.

3 N hc663 10

t

ATOMIC PHYSICS


3 3N 663 10 663 10hc 1242 nmeVt

540

9n / t 1N / t 5 10

3

9 9n 1 N 1 663 10 540t 5 10 t 5 10 1242 nmeV

11neI 5.76 10 A

t

Example 17. Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy whichrequires voltage V0 to prevent a electron from reading collector. In the same setup, light of wavelength 220nm, ejects electrons which require twich the voltage V0 stop them in reaching a collector. Find the numericalvalue of voltage V 0. (take planks constant, h = 6.6 10

34 Js and 1 eV = 1.6 1019 J)

Sol. 01

hc eV

and 02

hc 2eV

Here 1 = 330 nmand 2 = 220 nmAfter solving,

015V volt8

Example 18. A small 10W source of ultraviolet light of wavelength 99 nm is held at a distance 0.1 m from ametal surface. The radius of an atom of the metal is approximately 0.05 nm. Find

(i) the average number of photons striking an atom per second.(ii) the number of photoelectrons emitted per unit area per second if the efficiency of liberation of photoelectrons is

1%.

Sol. (i) I = intensity = 210

4 (0.1) w = power incident on atom

= 22 9210I r 0.05 10

4 10

n hcwt

n w 5t hc / 16

ATOMIC PHYSICS


(ii) no. of photons incident per unit area per secondI

hc /

no. of ejected electrons per unit area per second

20I 1 10hc / 100 80

Example 19. The surface of cesium is illuminated with monochromatic light ofvarious wavelengths and the stopping potentials for the wavelengths aremeasured. The results of this experiment is plotted as shown in the figure.estimate the value of work function of the cesium and Plancks constant.

12

012

0.49

0.5 1.0 1.5 v 10 Hz15

supp

ortin

g po

tent

ial (

volt)

Sol. h = + eVs

or shVe e

From graph

15h 2e 0.49 10

19

152h 1.6 10

0.49 10

h = 6.53 1034 Js

But 2e

= 2 eV

Example 20. In a photoelectric effect set-up a point source of light of power 3.2 103 W emits monoenergeticphotons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationarymetallic sphere of work function 3.0 eV and of radius 8.0 103 m. The efficiency of photoelectronemission is one for every 106 incident photons. Assume that the sphere is isolated and initially neutral andthat photoelectrons are instantly swept away after emission.

(a) Calculate the number of photoelectrons emitted per second.(b) Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectrons

emitted.(c) It is observed that the photoelectron emission stops at a certain time t after the light source is switched on

why?(d) Evaluate the time t.Sol. (a) Energy of emitted photons

E1 = 5.0 eV = 5.0 1.6 1019 J

E1 = 8.0 1019 J

Power of the point source is 3.2 103 watt or 3.2 103 J/sTherefore, energy emitted per second,

E2 = 3.2 103 J.

s 0.8 m r = 8.010 m3

Hence number of photons emitted per second

ATOMIC PHYSICS


21

1

En

E or

3

1 193.2 10

n8.0 10

n1 = 4.0 1015 photons/sec.

Number of photons incident on unit area at a distance of 8.0 m from the source S will be

n2 = 1

2n

4 (0.8) = 154.0 10

4 (0.64)

5.0 1014 photon/sec m2.

The area of metallic sphere over which photons will fall is :A = r2 = (8 103)2 m2 2.01 104 m2

Therefore, number of photons incident on the sphere per second aren3 = n2 A = (5.0 10

14 2.01 104) 10111 per secondBut since one photoelectron is emitted for every 106 photons hence number of photoelectrons emitted persecond,

n = 36n

10 = 11

61010

= 105 per second

or n = 105 per second(b) Maximum kinetic energy of photoelectrons

Kmax = Energy of incident photones work functionKmax = (5.0 3.0) eV = 2.0 eV = 2.0 1.6 10

19 JKmax = 3.2 10

19 JThe de-Broglie wavelength of these photoelectrons will be

1max

h hp 2 K m

(Here h = Plancks constant and m = mass of electron)

34

1 19 31

6.63 102 3.2 10 9.1 10

1 = 8.6810

10 m = 8.68

Wavelength of incident light 2 (in ) = 1

12375E (in eV)

or 2 = 12375

5 = 2476 Therefore, the desired ratio is

2

1

2475 285.18.68

(c) As soon as electrons are emitted from the metal sphere, it gets positively charged and acquires positivepotential. The positive potential gradually increases as more and more photoelectrons are emitted from itssurface. Emission of photoelectrons is stopped when its potential is equal to the stopping potential requiredfor fastest moving electrons.(b) As discussed in part (c), emission of photoelectrons is stopped when potential on the metal sphere isequal to the stoppeing potential of fastest moving electrons.Since Kmax = 2.0 eV

ATOMIC PHYSICS


Therefore, stopping potential V0 = 2 volt. Let q be the charge required for the potential on the sphere to beequal to stopping potential or 2 volt. Then

9 30

1 q 92 . 9.0 104 r 8.0 10

q = 1.78 1012 CPhotoelectrons emitted per second = 105 [part a]or charge emitted per second = (1.6 1019) 105 C

= (1.6 1014) CTherefore, time required to acquire to charge q will be

2

14q 1.728 10

t sec sec1.6 10 1.6

or t 111 second

Example 21. Monochromatic radiation of wavelength 1 = 3000 falls on a photocell operating in saturationmode. The corresponding spectral sensitivity of photocell is J = 4.8 mA/W. When another monochromaticradiation of wavelength 2 = 1650 and power P = 5 mW is incident. It is found that maximum velocity ofphotoelectrons increases to n = 2 times. Assuming efficiency of photo-electron generation per incidentphoton to be same for both the cases, calculate(i) threshold wavelength for the cell and(ii) saturation current in second case.[Given, h = 6.6 1034 Js, c = 3 108 ms1 and e = 1.6 1019 coul.]

Sol. Maximum kinetic energy of photoelectrons is given by Ek = hc

W0, where is wavelength of incident

radiation and W0 is work function of the surface on which radiation is incident. Maximum kinetic energy of photoelectrons emitted by radiation of wavelength 1 is

given by 12 mv1

2 = 1

hc

W0

or21 0

1

hcmv 2 W

...(1)

where m is mass of an electron and v1 is maximum velocity of photoelectrons.

Similarly, for radiation of wavelength 1, 22 0

2

1 hcmv W2

...(2)

But v2 = 2v1, therefore from equation (2),

21 0

2

hc2mv W ...(3)

From equations (1) and (3),

0 01 2

hc hc4 W W

or W0 = 3 eV

ATOMIC PHYSICS


But work-function W0 = 0

hc where 0 is threshold wavelength.

0 = 0

hcW = 4125 Ans. (i)

In saturation mode, spectal sensitivity with wavelength 1 = 3000 is J = 4.8 mA/W or 4.8 mC/J. It meanswhen 1 joule radiation of wavelength 1 = 3000 is incident, a charge of 4.8 mC flows in saturation mode

or 4.8 mC

e electrons are ejected.

Energy of each photon of wavelength 1 is E1 = 1

hc

Number of photons in 1 joule radiation of wavelength 11

1

1E hc

No. of electrons ejected by these photons = Je =

3

194.8 101.6 10

= 3 1016

Efficiency of photo-electron generation per incident photon,16

1

3 100.0198

( / hc)

.

Energy of each photon of wavelength 2,

E2 = 2

hc

Rate of incidence of photons of wavelength 2 in a radiation of power P

2

2

PPE hc

per second

Since, efficiency of photo-electron generation is same for both the case, therefore, rate of ejection ofelectrons in later case

2P.hc

per second

Rate of flow of charge in saturation mode = 2Phc

e Cs1 = 13.2 Cs1

But rate of flow of charge is current. Hence, saturation current is second case = 13.2 A. Ans. (ii)

Example 22. A monochromatic point source S radiating wavelength = 6000 with power P = 2 watt, an aperture A of radius R = 1 cmand a large screen are placed as shown in fig. A photoemissive detectorD of surface area S = 0.5 cm2 is placed at centre of the screen.Efficiency of detector for photoelectric emission per incident photonis = 0.9.

D

A

LS

60cm

6 m(i) calculate photon flux at centre of screen and photo current in the detector.(ii) If a convex lens L of focal length f = 30 cm is inserted in the aperture as shown, calculate new value of

photon flux and photo current assuming a uniform average transmission of 80% from the lens.(iii) If work function of photo-emissive surface is W0 = 1 eV, calculate value of stopping potential in two cases

(without and with the lens in aperture). Given, h = 6.625 1034 J-S, c = 3 108 ms1.

ATOMIC PHYSICS


Sol. Photon flux is rate of incidence of photons per unit are of detector. Therefore, to calculate photon flux, firstrate of emission of electrons from the source should be calculated.

Energy of each photon ishcE

Rate of emission of energy from the source is P = 2 watt = 2 Js1

Rate of emission of photons from the source is P PnE hc

or n = 6.04 1018 photons per second(i) Distance of detector from source is r1 = 6 m

Photon flux at detector,

161 2

1

n 1.33 104 r

photons/m2s

Rate of incidence of photons on detector = 1 . SRate of emission of electrons from detector = 1S per secondSince, current is charge flowing per second, therefore photo current

= (1S)e = 9.6 108 amp

(ii) When a concave lens is inserted in the aperture, it refracts incident rays. Therefore, photon flux and hencephoto-current changes.Distance of lens from source is r2 = 0.60 m

Photon flux at lens is = 22

n4 r = 1.33 10

18 photons/m2s

Considering a very small area A of the lens,Rate of incidence of photons on this area of lens = ANow considering refraction through the lens,

u = 60 cm, f = + 30 cm v = ?

Using lens formule,1 1 1v u f , v = + 60 cm

Since, average transmission from lens is 80%, therefore, rate of transmission of photons from area A of lens= 0.8 A

But these photons are transmitted in a solid angle subtended by the area A at P as shown in fig.

This solid angle, 2 2A Av (0.6)

Rate of photons transmitted per unit solid angle is(0.8 A )T

= (0.8 ) (0.6)2 = 0.288

DPS

4.80mv

ATOMIC PHYSICS


Solid angle subtended by unit area of detector at P, = 21

(4.80) Photon flux at D, 2 = T = 1.67 10

16 photons/m2sRate of incidence of photons on detector = 2SRate of emission of electrons from detector = 2S

Photo current = (2S)e = 1.20 107 amp

(iii) Since, stopping potential V0 is given by e.V0 = 0hc W and and W0 both remain unchanged, therefore,

stopping potential is same for both the cases.

0 01 hcV W 1.07 volte

PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHTAccording to wave theory, when light falls on a metal surface, energy is continuously distributed over thesurface, energy is continuously distributed over the surface. All the free electrons at the surface receive lightenergy. An electron may be ejected only when it acquired energy more than the work function. If we use alow-intensity source, it may take hours before an electron acquires this much energy from the light. In thisperiod, there will be many collisions and any extra energy accumulated so far will be shared with theremaining metal. This will result in no photoelectron. This is contrary to experimental observations. Nomatter how small is the intensity, photoelectrons are ejected and that too without any appreciable timedelay. In the photon theory, low intensity means less number of photons and hence less number of electronsget a change to absorb energy. But any fortunate electron on which a photon falls, gets the full energy of thephoton and may come out immediately.In figure, we illustrate an analogy to the wave the particle behaviour of light. In part (a), water is sprayedfrom a distance on an area containing several plants. Each plant receives water at nearly the same rate. Ittakes time for a particular plant to receive a certain amount of water. In part (b) of the figure, water is filledin identical, loosely-tied water bags and a particle physicist throws the bags randomly at the plants. When abag collides with a plant, it sprays all its water on that plant in a very short time. In the same way, whole ofthe energy associated with a photon is absorbed by a free electron when the photon hits it.

(a) (b)The maximum kinetic energy of a photoelectron does not depend on the intensity of the incident light. Thisfact is also not understood by the wave theory. According to this theory, more intensity means more energyand the maximum kinetic energy must increase with the increase in intensity which is not true. The dependenceof maximum kinetic energy on wavelength is also against the wave theory. There should not be any thresholdwavelength according to the wave theory. According to this theory, by using sufficiently intense light of anywavelength, an electron may be given the required amount of energy to come out. Experiments, however,show the existence of threshold wavelength.

ATOMIC PHYSICS


DUAL NATURE OF LIGHT(a) Wave nature:

Wave nature of light can be explained on the basis of reflection, refraction, interference, diffraction andpolarization.

(b) Particle Nature:Energy is transported by energy particles, photons. It could be explained by photoelectric effect, Zeemaneffect, Compton effect etc.

MATTER WAVE THEORY OR DE-BROGLIES THEORY(a) This theory was given on the basis of duel nature of light.(b) According to de-Broglie theory each and every moving particle has some wave nature associated with itself

which is called matter waves.(c) Thus, moving particles like e, proton, neutron, -particle etc, also behave like waves.(d) These waves are waves of probability.(e) The wavelength associated with a moving particles is given by = h/p, where p is the momentum of the

particle.(f) This wavelength is known as the de-Broglie wavelength of the particle.

DE BROGLIE WAVELENGTHS

For Photon For moving particle

Rest mass Zero m

Effective mass m = 2Ec = 2

hc

m = 02 2

m1 v / c

Energy (kinetic) E = hv = hc

E = 12 mv

2, E = 2p

2m

Momentum p = Ec =

hc

= h

p = mv = 2mE

Wavelength = hp =

hcE =

hp

= h

2mE

Wavelength for charged particle = h

2mEaccelerated by V volts K.E. = qV

Speed c = 3 108 m/s v = 2E / m = 2qV / m

De-Broglies explanation for stable Bohrs orbits :(a) De-Broglie suggested that non-radiation of energy by the electrons circling

in a Bohrs orbit can be explained on the basis on the basis of the formationof stationary waves by the electrons in circular motion in Bohrs orbits.

rO

Fourth Bohr-orbit

(b) Comparing this to the vibrations of a wire loop such stationary waves wouldbe formed if each wave joins smoothly with the next.

ATOMIC PHYSICS


(c) In other words the number of wavelengths must be an integer.(d) Condition for stable orbits :

An election can circle around an atomic nucleus without radiation energy if the circumference of its orbit isan integral multiple of the electrons wavelength.

i.e. 2r = n condition for stable orbits.

2r = nhp pr =

nh2 mvr =

nh2 [ p = mv]

mvr = nh2

As mvr is the angular momentum of the circling electron. Bohrs postulate is justified.

C18: Calculate the de-Broglie wavelength associated with the motion of earth (mass = 6 1024kg) orbitingaround the sun at a speed of 3 106 ms1.

Sol: = h

mv = 34

24 6 1

6.63 10 (Js)(6 10 ) (3 10 )(kg ms )

= 3.68 1065 m

NOTE: The wavelengths associated with the motion of macroscopic objects like earth, train etc, arenegligible compared to their sizes. This is why the wave-like character of these objects is not

observable in our daily life.

C19: Calculate the de-Broglie wavelength of an -particle of mass 6.576 1027 kg and charge 3.2 1019coulomb, accelerated though 2000 V.

Sol: E = kinetic energy of a-particle = qV E = 3.2 1019 2000 J = 6.4 1016 J

de Broglie wavelength,

= h

2mE, [ 2mE = momentum of photon]

= 34

27 16

6.63 10 (J s)2 6.576 10 6.4 10 (J kg)

= 2.28 1013 m

NOTE: The wavelength associated with -particles is of the order of size of the -particle.That is why the wave like character of a-particle is observable.

C20: A particle of mass m and charge q is accelerated through a potential difference V. Find (a) its kinetic energy(b) momentum, and (c) de-Broglie wavelength associated with its motion.

Sol: When the particle is accelerated through a potential difference V, gain in kinetic energy is given by K = qV.(a) Thus, kinetic energy, K = qV.

(b) Momentum of the particle (p) is given by K = 2p

2mp = 2mK = 2mqv

Momentum = 2mqv

ATOMIC PHYSICS


(c) De- Broglie wavelength is given by,

= hp =

h2mqV

Thus, wavelength = h

2mqV

Example 23. Assume that the de-Broglie was associated with an electron can form a standing wave betweenthe atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that onesuch standing wave is formed if the distance d between the atoms of the array is 2 . A similar standingwave is again formed if d is increased to 2.5 but not for any intermediate value of d. Find the energy ofthe electron in eV and the least value of d for which the standing wave of the type described above canform.

Sol. From the figure it is clear thatp . (/2) = 2 (p + 1) . /2 = 2.5

/2 = (2.5 2.0) = 0.5 or = 1 = 1010 m(i) de Broglie wavelength is given by

2

2.5

N Np-loops

(p + 1) loops

/2

h hp 2 km

K = kinetic energy of electron2 34 2

172 31 10 2

h (6.63 10 )K 2.415 10 J2m 2(9.1 10 )(10 )

17

192.415 10K eV1.6 10

K = 150.8 eV(ii) N N The least value of d will be when only one loop is formed

dmin = /2 or dmin = 0.5

C21: Find the de-broglie wavelength associated with an electron accelerated through a potential difference of30 kV.

Sol: Kinetic energy of the electronK = qV = e(30 kV) = 30 keV = 30 103 1.6 1019 J = 4.8 1015 J

Now, momentum of the electron = 2mK = 34 152 9.1 10 4.8 10 kg.J = 2.96 1024 kg.m/s

De-broglie wavelength, = h

Momentum

=34

24

6.63 10 J s2.96 10 Kg m / s

= 2.24 1010 m = 2.24 A

ATOMIC PHYSICS


ATOMIC STRUCTURE

ATOMIC MODELBy now, it is well-known that matter, electricity and radiation etc. are all atomic in character. Although, noone has so far seen individual atoms, there is no doubt that they really exist. In 1895, it was discovered byJ.Perrin in Paris that the cathode rays consist of negatively-charged practices called electrons. In 1897,J.J. Thomson measured the e/m ratio for an electron whereas its charge was measured by Millikan in1906 by his famous oil-drop experiment. Mass of the electron was found by dividing charge e by the ratioe/m. Discovery of positive rays during the latter part of 19th century indicated that a normal atom consistedof both negative and positive charges. But how these charges are distributed in an atom was not known atthat time. Consinuous efforts have been made since then to study the physical structure of an atom such asits extra-nuclear electronic structure chiefly with the help of spectral properties of atoms. To account for thespectroscopic data obtained experimentally over the years, several theories regarding atomic structure havebeen proposed from time to time which are called the atomic models. Various atomic models proposed byscientists over the last few decades are:(i) Thomsons Plum pudding model, (ii) Rutherfords Nuclear model, (iii) Bohrs model (iv) SommerfeldsRelativistic model (v) Vector model and finally (vi) Wave-mechanical model.These different models have been suggested one after the other in an effort to get a satisfactory interpretationof the experimental data which, it is hoped, will ultimately lead to a perfect and complete understanding ofthe physical structure of an atom.

Thomsons Plum Pudding ModelAccording to this model, the atom is regarded as a heavy sphere of positivecharge seasoned with enough electron plums to make it electrically neutral.Thomson visualized the positive charge of an atom being spread out uniformlythroughout a sphere of about 1010 metre radius with electrons as smallerparticles distributed in circular shells as shown in figure. Whereas the net force + + +

+ +

+

+

+ +

+

+ +

+ + + + + + + + + + + + + ++ + +

+

+ +

+

+ +

+ + + + + + + + + + + + +

+ + +

+

+ +

+

+ + + + + + + + + + +

+ ++ +

+

+ +

+ + + + + + + +

+ +

+

+ + + + + +

+ + + +

exerted by the positively-charged sphere on each electron is towards the centre of the sphere, the differentelectrons experience mutual repulsion and get arranged in the form of circular shells.This atomic model was given up after some time because it could not provide any satisfactory mechanismfor explaining the large deflection suffered by -particles in Rutherfords experiment.

Rutherfords Experiment on -particle ScatteringAs shown in figure high-speed a-particle (i.e. helium nuclei each with a charge of +2e) from some radioactivematerial like radium or radon, confined to a narrow beam by a hole in a lead block, were made to strike avery thin gold foil G. While most of the -particles went straight through the foil as if there were nothing there(and produced scintillations on a fluorescent screen), some of them collided with the atoms of the foil andwere scattered around at various angles -a few being turned back towards the source itself.

150 4560

Radium

Lead block +

Beam

of-p

artic

les

ATOMIC PHYSICS


Although some small angle scattering could be expected from Thomsons model, large-angle scattering wasabsolutely not expected at all. Further detailed experiments in Rutherfords laboratory by Geiger and Marsdenshowed conslusively that large-angle scattering can be expected only if one assumes that a massive positivepoint charge exists at the centre of each gold atom as shown in figure.

According to this model proposed by Rutherford in 1911, the positive massive part of an atom is assumedto be concentrated in a very small volume at its centre. This central core, now called nucleus, is surroundedby a cloud of electrons which makes the entire atom electrically neutral.

P

+Ze

Nucleus

Asymptoteof trajectory

Trajectoryof -particle

The large-angle scattering of positively-charged -particlescould be easily explained on this atomic model as shown infigure. This scattering is due to the mutual repulsion (as perCoulombs law) between the -particles and the concentratedpositive charge on the nucleus. The -particle approachesthe positively-charged nucleus along AO. If there were norepulsion from the nucleus, it would have passed at a distanceof p from it. However, due tocoulombic force of repulsion, the -particle follows a hyperbola with nucleus as its focus. The lines AO andAO are the asymptotes of the hyperbola and present approximately the initial and final directions of the -particle when it has passed out of the effective range of the nuclear electric field.

As seen, the -particle is deflected through an angle . The perpendicular distance from the nucleus to theline AO is called the impact parameter and is denoted by p. The Rutherfords scattering formula is

tan 2

= 1 2Q Q2pE

where Q1 = Charge of the incoming -particleQ2 = Charge of the scattering nucleusE = Kinetic energy of the incident -particlep = Impact parameter

Distance of Closest ApproachSuppose that an -particle approaches a positively-charged nucleus for a head on collision with a kineticenergy of K. As shown in figure at point A, the repulsive force of the nucleus is so strong as to stop the -particle momentarily. At this point, all the kinetic energy of the -particle is converted into potential energy.Let D be the distance of closest approach of the -particle. The potential at point A due to nuclear chargeZe is

= 0

Ze4 D

+D

A

Potential energy of the -particle when at point A is

= 0

Ze.2e4 D =

2

0

2Ze4 D

K = 2

0

2Ze4 D or D =

2

0

2Ze4 K

Since -particles are generally obtained from natural radioactive substances, their kinetic energy K isknown. Hence, value of D can be found easily from the above equation.

ATOMIC PHYSICS


C22: In Geiger-Marsden experiment on -particle scattering from gold foil, the kinetic energy of -particlesused was 7.68 MeV. Calculate the distance of closest approach of -particle if atomic number of gold in79.

Sol: D = 2

0

2Ze4 K

Here, Z = 79, e = 1.6 1019 C; 0 = 8.854 1012 F/m

K = 7.68 MeV = 7.68 1.6 1013 J

D = 19 2

12 13

2 79 (1.6 10 )4 8.854 10 7.68 1.6 10

= 2.96 1014 m

Major Deficiencies in Rutherfords Nuclear ModelIt was found later on that Rutherfords model had two serious drawbacks concerning

(i) distribution of electrons outside the nucleus and(ii) the stability of the atom as a whole.

It can be shown that electrostatic forces between the positive nucleus and the static negative extra-nuclearelectrons are not enough to produce equilibrium in such a nuclear atom. For example, consider the case ofan atom* having two electrons and a nucleus with a charge of +2e. Suppose the electrons are symmetricallyplaced at a distance of r from the nucleus and are stationary. The force of attraction between the nucleus and

each of the electrons is F = e. 22e

4 r = 2

2

2e4 r

while the force of repulsion between the two electrons is

2

2

e4 4r

= 2

2

e16 r

. Since the force of attraction is eight times the force of repulsion, the electrons will

fall into the nucleus thereby destroying the stable structure of the atom.

+ +

+ +2e

+ +

+ +2e

v

vr

m

e+

Photon

(a) (b) (c)

2e

(h )v

To overcome this difficulty, Rutherford suggested that stability can be achieved (as in a solar system) byassuming that electrons, instead of being static, revolve round the nucleus with such a speed that the centrifugalforce balances the attractive force exerted by the nucleus on the electrons. As seen from figure (b), conditionfor stability is achieved when

2mvr

= 20

2e.e4 r or mv

2r = 2

0

2e4

In general, if Z is the atomic number, then nuclear charge is Ze, so that the above relation becomes2mv

r = 2

0

Ze.e4 r or mv

2r = 2

0

Ze4

ATOMIC PHYSICS


Incidentally, it may be noted that according to the above relation, it is possible to have an infinite number oforbits in which electrons can rotate.

But this assumption of revolving electrons led to serious difficulty from the point of view o the electromagnetictheory according to which an accelerated charge must continuously emit electromagnetic radiation or energy.Since an electron revolving in a circular orbit has centripetal acceleration (= v2/r), it must radiate energy asper the laws of classical electrodynamics.

Due to this continuous loss of energy, the electrons will gradually approach the nucleus by a spiral path andfinally fall into it as shown in figure (c). Hence, it is seen that the orbital motion of the electron destroys thevery purpose for which it was postulated i.e. the stability of the atom. Obviously, either Rutherfords nuclearatomic model with revolving electrons is defective or the classical electromagnetic theory fails in this particularcase. This dilemma was solved in 1913 by Neils Bohr who proposed an improved version of Rutherfordsatomic model.

Bohrs Atomic modelThis model (first proposed for hydrogen atom but later applied to other atoms as well) retains the twoessential features of Rutherfords planetary model i.e.

(i) the atom has a massive positively-charged nucleus and(ii) the electrons revolve round their nucleus in circular orbits the centrifugal force being balanced, as before, by

the electrostatic pull between the nucleus and electrons.However, he extended this model further by utilizing Plancks Quantum Theory. He made the followingthree assumptions:

(iii) an electron cannot revolve round the nucleus in any arbitrary orbit but in just certain definite and discreteorbits. Only those orbits are possible (or permitted) for which the orbital angular momentum (i.e. moment of

momentum) of the electron is equal to an integral multiple of h

2 i.e. orbital angular momentum = nh2 where

n is an integer and h is Plancks constant. Such orbits are also known as stationary orbits.(iv) while revolving in these permitted stationary (or stable) orbits, the electron does not radiate out any

electromagnetic energy. In other words, the permissible orbits are non-radiating paths of the electron.(v) the atom radiates out energy only when an electron jumps from one orbit to another. I f E2 and E1 are the

energies corresponding to two orbits before and after the jump, the frequency of the emitted photon is givenby the relation

E2 E1 = hv or E = hvwhere v is the frequency of the emitted radiations.

C 23: If I is the moment of inertia of an electron and its angular velocity, then as per assumption (iii) givenabove

I = nh2 or (mr

2) = nh2

or2(mr )v

r =

n.h2 or mvr =

n.h2

Alternatively, since the momentum of the revolving electron is mv, its momentabout the nucleus is = mvr

Hence mvr = n.h2 .....(i)

e

m

v

+zer

ATOMIC PHYSICS


where, n = 1, 2, 3 for the first second and third orbits respectively. It is called the principal quantum numberand because it can take whole number values only, it fixes the sizes of the allowed orbits (also called Bohrscircular orbits).

r1 r2 r3

n = 1n = 2n = 3

+Ze

E1E2E3

Permitted orbits(a)

KLM

+Ze

E1

Electron Jump(b)

EE2E3

Let the different permitted orbit have energies of E1, E2, E3 etc. as shown in figure (a). The electron can beraised from n = 1 orbit to any other higher orbit if it is given proper amount of energy. When it drops backto n = 1 orbit after a short interval of time, it gives out the energy difference E in the form of a radiation asshown in figure (b). The relation between the energy released and frequency of the emitted radiation is

E2 E1 = hv or E = hv

Expressions for velocity, radius, energy of electron and orbital frequency in Bohrs orbitHere it should be kept in mind that Bohrs model is valid only for hydrogen atom and hydrogen-like ions. Inother words, we can say that is applicable to hydrogen atom and ions having just one electron. Examples ofsuch ions are He+, Li++, Be+++ etc.According to Bohrs

+ze

e, mrvFirst postulate,

2

20

Ze4 r =

2mvr

where 0

14 = 9 10

9 Nm2 C2 v = velocity of electron

and 0 = 8.854 1012 C2/Nm2 m = mass of an electron

= Absolute permittivity Z = atomic numberof vacuum or free space r = radius of the orbit

e = magnitude of charge on an electronn = principal quantum number i.e. orbit number

r = 2

20

Ze4 mv .....(i)

From fourth postulate of Bohr, we have

mvr = nh2 .....(ii)

where, n = 1, 2, 3, 4, ........ i.e., n is a positive integer.From equations (i) and (ii), we get

mv 2

20

Ze4 mv =

nh2 v =

2

0

ze2 nh

Now, let us substitute the value of v in equation (ii),

ATOMIC PHYSICS


m 2

0

ze r2 nh

=

nh2

2 20

2

n hrmZe

Note that for fixed n, v Z, r 1Z

For fixed z, v 1n , r n

2

If V1 is the speed of the electron in the 1st orbit

Then, Vn = V1Zn , where V1 =

c137 , where c is speed of light.

If a0 = first Bohr radius = 0.53 , then

rn = a02n

Z

K.E. of the electron = 12 mv

2 = 2 4

2 2 20

mZ e8 h n = K =

2kZe2r

,

Potential energy of the atom = 2

0

Ze4 r

= U = 2kZe

r[k = 1/40]

U = 2 4

2 2 20

mZ e4 h n

Total energy of the atom, E = K + U

E = 2 4

2 2 20

mZ e8 h n

In general, En = 13.6 2

2

Zn

in eV..

Orbital frequency for the electron,

v = V

2 r = 2 2

2 20 0

Ze mZe2 2 hn h n

v =

2 4

2 3 30

mZ e4 n h

Time period of revolution (T) is given by

T = T1 3

2

nZ

where T1 = time period of revolution in the 1st orbit = 1.52 1016 s =

0

1

2 av

NOTE : It is assumed that the acceleration of the nucleus is negligible on account of its large mass.

Some important results for H-atoms when n = 11. Bohr radius, a0 = 0.53

rn = a0n2/Z

2. v1 = 2.18 106 ms1 c/137

3. E1 = 13.6 eV = 1U

2

ATOMIC PHYSICS


4. K1= 13.6 eV = 1U

25. U1 = 27.2 eV6. v1 = 6.6 10

15 Hz7. 1 rydberg = 13.6 eV8. B1 = 12.5 tesla, where B1 is the magnetic field at the centre of Bohr atom due to the current generated by

the motion of electron in 1st orbit.

Ground state and excited statesThe state of an atom with the lowest energy is called its ground state or normal state. For ground state, n =1.The states with higher energies are called excited states. For the first excited state, n = 2; for the 2nd excitedstate, n = 3 and so on.For mth excited state, n = m + 1

Ionization energy and ionization potentialThe minimum energy needed to ionize an atom is called ionization energy.The potential difference through which an electron should be accelerated to acquire the value of ionizationenergy is called ionization potential.The value of ionization energy of H-atom in ground state is 13.6 eV that of ionization potential is 13.6 eV.

Binding energyBinding energy of a system is the energy needed to separate its constituents to large distances or it may bedefined as the energy released when its constituents are brought from infinity to form the system.The value of binding energy of H-atom is 13.6 eV, identical to its ionization energy.

Excitation energy and excitation potentialThe energy needed to take the atom from its ground state to an excited state is called the excitation energyof that excited state.The potential difference through which an electron should be accelerated to acquire the value of excitationenergy is called excitation potential.

NOTE : (A) I.E. = E E1 = E1 = Binding energy of the H-atom.

(B) Movement of electron in circular orbits in a Bohr atom causes electric current in the orbit. Thiscurrent will lead to self-generated magnetic field in the atom and also magnetic current (m)

(a) Magnetic field(B)If B is the magnetic field generated at the centre of atom, then

H-atom+e

a0e

iB =

0

0

i2a

, where i is the current due to motion of the electron

and a0 is the 1st Bohr-radius.Hence, i = ev

hence B = 0

0

e2a

ATOMIC PHYSICS


Putting the values of v and a0, we get B = 12.5 Tesla(b) Magnetic movement vector ( )

= iA

= iA = ev 20a

(c) Relation between & L

(angular momentum vector) = iA = ev 20a , [i = ev, A =

20a ]

L = mV1a0 = m2v 20a [V1 = 2va0]

L

= 20

20

e am2 a =

e2m L

=

e2m

Vectorially,

=

e2m

L

Example 24: The quantum number of Bohr orbit in H-atom whose radius is 0.01 mm is(a) 223 (b) 435 (c) 891 (d) none of these

Sol: (b) We know thatrn = a0n

2

n2 =n

0

ra

n =3

10

0.01 100.529 10

n = 435

Example 25: The quantum number n in the Bohrs model of H-atom specifies:(a) radius of the electron (b) energy of the electron(c) angular momentum of the electron (d) all of these

Sol: (d) rn =2

02

hme

n2

En = 4

2 2 20

me8 h n

, Ln = nh

2

Example 26: The radius Bohrs orbit in ground state for H-atom is(Take 0 = 8.86 10

12 C2/Nm2, h = 6.6 1034 J-s)(a) 0.528 (b) 0.0528 (c) 5.28 1010 m (d) 5.28 1010 cm

Sol: (a) a0 =2

02

hme

(in this case, Z = 1, n = 1)

a0 =34 2 12

31 19 2

(6.6 10 ) (8.86 10 )3.14 (9.1 10 ) (1.6 10 )

= 0.528

ATOMIC PHYSICS


Example 27: The speed of the electron in the first Bohr orbit of H-atom is (Take c = speed of light in vacuum)

(a) c (b) c

13.6 (c) c

137 (d) 137 cSol: (c) For H-atom, we know that

v =2

0

e2 h [since n this case, Z = 1 and n = 1]

=2

0

e c2 hc

But 2

0

e2 hc =

1137 = fine structure constant

v =c

137

Example 28: In a H-atom, binding energy of the electron in the ground state is E1. Then the frequency ofrevolution of the electron in the nth orbit is

(a) 132En h (b)

312E nh

(c) 132mEn h

(d) none of these

Sol: (b) The frequency of revolution of the electron in the nth orbit is given by

v = 2 4

2 3 30

mZ e4 h n =

13

2Ehn where E1 =

2 4

2 20

mZ e8 h

C24: Calculate the energy of a He+ ion in its first excited state.

Sol: En = 2

2

(13.6eV)Zn

Here, Z = 2, n = 2

En = 2

2

13.6 2n

eV = 13.6 eV

C25. An electron in a hydrogen like atom is in an excited state. It has a total energy of 3.4 eV. Calculate :(i) the kinetic energy(ii) the de-Broglie wavelength of the electron

Sol. (i) Kinetic energy of electron in the orbits of hydrogen and hydrogen like atoms= | Total energy |

Kinetic energy = 3.4 eV(ii) The de Broglie wavelength is given by

h hP 2 Km

(K = kinetic energy of electron)

Substituting the values, we have34

19 31

(6.6 10 J s)2(3.4 1.6 10 J)(9.1 10 kg)

= 6.63 1010m or = 6.63

ATOMIC PHYSICS


Example 30. The electron in a hydrogen atom makes transition from M shell to L. The ratio of magnitudes ofinitial to final centripetal acceleration of the electron is(A) 9 : 4 (B) 81 : 16 (C) 4 : 9 (D) 16 : 81

Sol. rn n2

and vn 1n

2n

nn

va

r

n 2 21a

n n

n 41an

n 4Kan

M 3Ka a81

{ M = 3}

L 2Ka a16

{ L = 2}

M

L

a 16a 81

Hence (D) is correct.

Example 31. The angular momentum opf an electron in the hydrogen atom is 3h2 . Here h is Plancks constant.

The kinetic energy of this electron is :(A) 4.53 eV (B) 1.51 eV (C) 3.4 eV (D) 6.8 eV

Sol.nh 3hL2 2

n = 3In the electronic third orbit, the energy of electron

2

2Z

E 13.6n

1E 13.69

13.6E 1.51 eV9

Hence (B) is correct.

Example 32. A particle of mass m moves along a circular orbit in a centrosymmetrical potential field

U(r) = 2kr

2. Using the Bohrs quantization condition, find the permissible orbital radii and energy levels of

that particle.

ATOMIC PHYSICS


Sol. F = dUdr = kr

Negative sign implies that force is acting towards centre. The necessary centripetal force to the particle isbeing provided by this force F. Hence

2mvr

= kr ...(1)

and mvr = n hhwhere h

2 ...(2)

solving equations (1) and (2) we get

r = rn = n hm where =

km

and total energy E = U + K = 2kr

2 +

12 mv

2

Substituting the values, we getE = n h = En

Example 33. In a hypothetical system a particle of mass m and charge 3q is moving around a very heavyparticle having charge q. Assuming Bohrs model to be true to this system, the orbital velocity of mass mwhen it is nearest to heavy particle is

(A) 2

0

3q2 h (B)

2

0

3q4 h (C) 0

3q2 h (D) 0

3q4 h

Sol.2

emv

Fr

2

20

q 3q mv4 r r

2

0

3 qmvr v

4

hmvr

2

2

0

3 q hv

4 2

2

0

3 qv2 h

Hence (A) is correct.

ATOMIC PHYSICS


SPECTRUMDispersed light arranging itself in a pattern of different wavelength is referred to as a spectrum. Light comingfrom a source may be dispersed by a prism or by any other dispersing medium.When white light falls on a prism and the transmitted light is collected on a white wall or white paper then aspectrum is obtained which consists of different colours from red to violet.

Kinds of spectra :(A) Emission spectra: When a light beam emitted by certain source is dispersed to get the spectrum, it is

called an emission spectrum.

An emission spectrum can be three types :(a) Continuous spectrum: That emission spectrum which is obtained by continuously varying wavelength, is

called continuous emission spectrum. In this case, when light is dispersed, a bright spectrum continuouslydistributed on a dark background is obtained. Light emitted from an electric bulb, a candle or a red hot ironpiece comes under this category.

(b) Line spectrum: The atoms and molecules can have certainfixed energies. An atom or molecule, in an excited state, canemit light to lower its energy. Light emitted in such a processhas certain fixed wavelengths. When such a light is dispersed,certain sharp bright lines on a dark background is obtained.Such a spectrum is called line emission spectrum.

Atomic energy levels

Ener

gyLine spectrum

For example, when electric discharge is passed through sodium vapour, they vapour emits light of thewavelength 589.0 nm and 589.6 nm. When this light is dispersed by a high resolution grating , one obtainstwo bright yellow lines on a dark background.

(c) Band spectrum : The wavelengths emitted by the molecular energy levels which are generally grouped intoseveral bunches, are also grouped; each group being well separated from the other. The spectrum looks likeseparate bands of varying colours. Such a spectrum is called band emission spectrum.

Molecular energy levels

Ener

gy

Band spectrum

(B) Absorption spectrum : When white light is passed through an absorbing material, the material may absorbcertain wavelengths selectively. When the transmitted light is dispersed, dark lines or bands at the positionsof the missing (absorbed) wavelengths are obtained. Such type of spectrum is called absorption spectrum.

White light

Absorbing materialAbsorptionspectrum

ATOMIC PHYS

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