ATOMS AND NUCLEI

1. What is the value of ‘n’ for the upper most line and the lower most line in the energy level diagram?Ans : Upper most line, n=α and En = 0Lowermost line n= 1 and E1 = - 13.6ev.

2 . In Balmer series, to which level the transition from other orbits end up?

Ans : 2nd level.

3. What is the function of Breeder reactor?Ans : Converts fertile material to fissile material.

4 .What is MRI stand for? What is the principle of MRI ?

Ans : MRI –Magnetic Resonance Imaging. MRI is based on the principle of Nuclear Magnetic Resonance.

5 (a) The energy level diagram of an element is given below. Which of them will result in the transition of a photon of wave length 275nm?

(b) Which transition corresponds to emission of radiation of maximum wave length and minimum wave length?

__________________________________________________ 0ev ___A_______________________________________________ -2ev B

C _____________________________________________________ - 4.5ev

D _____________________________________________________ - 10ev

ANS: (a) 275nm corresponds to U V radiation

E= h c /λ and E= 6.63x10-34x3x108/275x10-9x1.6x10-9 = 4.5ev

This corresponds to’ B’.

(b) E= h c /λ => λ = h c /E

E α 1/λ Emission of radiation of maximum wavelength corresponds to “A” Emission of radiation of maximum wavelength corresponds to’D”

6. The energy levels of an element are given below:

_________________________________________________ -0.85ev

_________________________________________________ -1.5 ev

- 3.4 ev-

- -13.6 ev

Identify, using necessary calculations, the transition, which corresponds to the emission of a spectral line of wavelength 482 nm:ANS: As h f = E1 – E2, h c /λ = E1 – E2

λ= h c/ E1 – E2

λ= 6.626x10-34x3x108/2.55x1.6x10-9 =482nm.

7. How many α and β-particles are emitted when 90Th 232 changes to 82 Pb 208

90Th232 82 Pb208 +2 He4 + Y-1 e0

According to low of conservation of atomic number and mass number 90 = 82 + 2x – y2x – y = 8232 = 208 + 4x => x = 6.

Q 7: A radioactive isotope has a half-life of T years. How long will it take the activity to reduce

(a)3.125% (b)1% of its original value

Sol: = n …(1)

a) = = 5rom (1), (1/2)5 = (1/2)n

n=(5 half lives) , or =5

t=5T

b) = = (1/2)n

/ = = 6.64

t = 6.64T

Q 8: Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of Indus-Valley civilization. Given living plant gives about 15 decays per minute and half-life of carbon 14=5730 years.

Sol. Activity, R=

Initial activity, = N0

Given = =

From relation N=N0 , we have =

Or - = or

=

=

=4224.3 years

Q 9: Obtain the amount of Co necessary to provide radioactive source of 8.0mCi strength.

The half-life of Co is 5.3 years.

Avogadro Number = 6.02 1023 per g-atom.

Sol: We have R= …(1)

Given R=8.0mci = 8.0 10-3ci

= 8.0 10-3 3.7 10s-1= 29.6 107s-1

= = = s-1 =4.15 -9 s-1

From equation (1)

Number of undecayed nuclei, N= = = 7.13 1016

The mass of 6.02 1023 atoms is 60 grams, so mass of N=7.13 1016 atom is

=7.13 1016 ( grams

Required mass of Co, m= g = 7.1 10-6 g

Q 10: The half life of is 28 years. What is the disintegration rate of 15 mg of this

isotope?

Sol. We have

Disintegration constant

Here seconds

90g of contain 6.023 atoms

Number of Sr-90 atoms in 15 mg (=15 g)

N= x 6.023x

Disintegration rate, x1.00x =7.85x Bq

Q 11: The Q-value of a nuclear reaction

A + B C + D

Is defined by Q = ( mA + mB- mC- mD) c2

where the masses refer to the nuclear rest masses. Determine from the given data whether the following reactions are exothermic or endothermic.

(1) H + 1H3 1H2 +1H2

(2) 6C12 + 6C12 10Ne20 + 2He4

Atomic masses are given to be:mH = 1.007825um(1H2) = 2.014102um(1H3) = 3.016049um(6C12) = 12.000000um(10Ne20) = 19.992439um(2He4) = 4.002603uTake 1 u=931MeV

Sol: (1) Nuclear reaction is

P + 1H31H2 + 1H2 +Q

Mass of LHS= mH + m(1H3)=1.007825+3.016049 = 4.023874u

Mass of RHS= m(1H2) + m(1H2) = 2.014102+2.014102 = 4.028204u

Q= [( mA + mB- mC- mD)in Kg] c2 joule

= [( mA + mB- mC- mD)u] 931MeV

= {mH+ m(1H3)} – { m(1H2)+ m(1H2)} 931MeV

=[4.023874-4.028204] 931MeV = -0.00433 931MeV = -4.031MeV

As Q is negative, energy must be supplied for the reaction; hence the reaction is endothermic

(2) Nuclear reaction is 6C12 + 6C12 10Ne20 + 2He4

Q=m{(6C12) +m( 6C12 {m(10Ne20 ) + m(2He4)}

= (12.000000+12.000000) – (19.992439+4.002603)

= 24.000000 – 23.995042 = 0.004958u = 0.004958 931MeV

= 4.616 MeVAs Q is positive, the energy will be liberated in the reaction, hence the reaction is exothermic.

Q 12: How long an electrical lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as:

n +3.2MeV

Sol. Number of deuterium atoms in 2.0 kg is =6.02x

Number of reactions = = 3.01x

Energy released in one reaction = 3.2 MeV

Total energy released, W=3.01x x3.2MeV

=9.632x MeV

=9.632x x1.6x J=15.4x J

If t second is the required time during which the bulb glows, then

W=Pt gives

t= = =15.4x s

= years =4.9x years

Q 13: A radioactive material is reduced to of its original amount in 4 days. How much

material should one begin with so that 4 10-3 kg of material is left after 6 days?

Sol. = ( )n

Where n = is number of half lives.

Given = = ( )4

( )4 = ( ) n or n = 4

Given t = 4 days = 4

Half life, T = = = 1 day

If m0 is initial mass of radioactive material, then = ( )n.

Here n = = = 6, m = 4 10-3 kg = ( )6 =

Or m0 = 64m = 64 4 10-3 kg = 0.256 kg

Q14: Calculate the energy released if –nucleus emits an - particle.

OR

Calculate the energy released in MeV in the following nuclear reaction

U Th + He +Q

Given Atomic mass of 238U = 238.05079u

Atomic mass of of234Th = 234.04363

Atomic mass of alpha particle = 4.00260u

1u = 931.5 M e V/c2

Is the decay spontaneous, give reason.

Sol: The process is U Th + He +Q

The energy released

Q = (Mu – MTH – MHe)c2

= (238.05079 – 234.04363 – 4.00260)u c2 = (0.00456u) c2

= 0.00456 ) . c2= 4.25MeV

Q15: The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level – 0.85 eV to -3.4 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?

Sol : As we know:

En = eV ….(1)

For n = 1, E1 = -13.6 eV

When electron undergoes transition from EA = -0.85 eV to EB = -3.4eV

Then, from equation (1)

-0.85 = nA = 4

Similarly, -3.4 = nB = 2

Hence electron transits from n=4 to n=2. It corresponds to Balmer series.

We know, = R - )

Here, nA = 4, nB = 2, R = 1.097 107m-1

Then, = 1.097 107( - ) = 4862

16. An electron hydrogen atom the ground state energy is excited to

n= 4state. Calculate the energy absorbed and the wavelength of the electromagnetic radiation emitted when the atom comes to ground state.

Ground state energy of hydrogen atom= - 13.6 ev.

Ans: Ground state energy of hydrogen atom= - 13.6 ev. n= 4,

E4 =-13.6/n2 = -13.6 / 16 =- 0.8.5ev

Energy absorbed =-0.8.5ev –(-13.6ev) =+12.75ev

Wave length of radiation emitted when the atom comes to ground state is

= h c / (Eλ 4 – E1 ) e = 6.626x10-34x3x108/[-0.85-(-13.6)]x1.6x10-19

=0.97426x10-7m or = 974.26Aλ o

****************************************************************************************************************************************