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EXERCISE –I

1 . P(D) = 1 – e–t = 1 – e– × 2/= 2

2

e 1

e

2 . TA =

1

2hr, T

B =

1

4hr; T

A+B =

1 112 4 hr

1 1 6

2 4

so, first 1

2hr = 1 half lives (by A)

next 1 hr = 4 half lives (by B)

next 1

2 hr = 3 half lives (by A+B)

thus N = 08

N

2 ( Total eight half lives)

3 .dp

Fdt

(2P sin )

Ft

F = 2npsin [n = number of photon]

1N= 2(n) h

sin 30°n = 1027

4 . In photo electric effect the maximum velocity of e–

will corresponding to KEmax

& other are less thanit.

5 . KEmax

= h

6 . For threshold frequency, h0= 0 h

KEmax

= h–= h(–0)

saturationI n where I= nh

7 . Kmax

= h – Vs = maxK 4eV

e e 4 volts

8 .1 1 1

2 2 2

K h 1 0.5 1

K h 2.5 0.5 4

9 . 1 2

2 1

2.532948

5.06 5896

Å

1 0 . eVS = h

1 1 . Ek = h

1 2 . K.E.max

= frequency depends on

of light properties

of cathode

h

1 3 .1 2

2 1

1 4 .

0

0

0

hce 6V hc

4hce 2V 4

2

1 5 . I 2

1

r intensity becomes

1

4th

1 6 .6 10 01

2

2 0 0

2h hv 1v 8 10 ms

v 5h h 2

1 7 . M = M0e–t t0

0

MM e

20 n

1

20

= –t

1 / 2

n(20)T t

n2

1 / 2

n(2) n(10)T t

n2

t = 16.42 days

1 8 .2

0

hc 1mv

2

;

2

0 1

4hc 1mv

3 2

2 2

0 0 1

4 4 1 10 mv mv

3 3 2 2

2 2 01

1 4 1mv mv

2 3 2 3

1

4v v

3

1 9 . (a) P = N hc

(b)

0.1n N

100

(c) i n e

2 0 . 22 2

1

1 2

hchc2

K

hcK hc2

K1 < 2K

2

2 1 . Greater work function greater intercept

2 2 . De broglie waves are independent of shape & sizeof the object.

2 3 . De broglie waves are probability waves and areapplicable for all the objects. Wave nature isobserved for the small particles like electrons.

2 4 . K.E. q V

1

1 1 1

h h

p 2m K ; 2

2 2 2

h h

p 2m K

1 2q q q & V is same 1 2

2 1

m

m

UNIT # 12 (PART - I)

MODERN PHYSICS (Atomic and Nuclear physics)

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2 5 . = h

p(so same)

2 6 .h

2mK ;

3K kT

2

h h 25.15Å

3 3mkT T2m kT

2

2 7 . Maximum photon energy = 13.6 eV (emitted)So K

max = 13.6 – 4 = 9.6 eV

Hence stopping potential is – 9.6 VSo – 10V can stop

2 8 . No. of electron that can accommodate in nth shell = 2n2

Total number of elements= 2(1)2 + 2(2)2 + 2(3)2 + 2(4)2= 60

2 9 . E = 2

13.6eV

2= 3.4 eV

3 0 . E4 – E

3 = Z2

21 1 7Z

9 16 144

E2 – E

1 = Z2

21 31 Z

4 4

E4–E

2 =

2 21 1 3Z Z

4 16 16

3 1 . E = h 2 2

1 2

1 1

n n

max

for n2=2 to n

1=1

3 2 . Wave number7

2 2

1 1 11.097 10

1 n

Now in series limit corresponds to n = to n = 1

Wave number for series limit = 1

=1.097× 107

3 3 . E = 24.6 + 2

2

13.6(2)

1= 79.0 eV

3 4 . N =nC2 = 5C

2 = 10

3 5 . 2r = n()

3 6 . E = hc 1242eV nm

0.021nm

= 59 keV

3 7 . High atomic no. and high melting point.

3 8 .2

2 2

1 2

1 1 1R(Z 1)

n n

For K line, n

1 =1, n

2 = 2

22

2 2

1 1 1 3R(43 1) R 42

41 2

...(i)

and

22

2 2

1 1 1 3R(29 1) R 28

41 2

...(ii)

Dividing eq. (i) by (ii), we get 9 9

4 4

3 9 .1

= R(Z–1)2 2 2

1 2

1 1

n n

For wavelength of K, n

1 = 1 to n

2 = 2

1875 R

4= R (Z

A–1)2

11

4

...(i)

and 675 R

1= R[Z

B–1]2

11

4

...(ii)

By solving eq. (i) & (ii) we getZ

A = 26 and Z

B = 31

[Four elements lie between these two]

4 0 . min

1

V so

1

6.22 10

min = 0.622 Å

4 1 . Characteristic X–rays corresponds to the transitionof electrons from one shell to another.

4 3 . N x

(200 × 106 × 1.6 × 10–19) = 1000

4 4 . Q = –7(5.6) + [4(7.06)]2 Q = 17.3 MeV

4 5 . 42 He 2n 2p

BE = [2(1.0073 + 1.0087) – 4.0015] 931.5 MeV

4 6 . 2 Deuteron 42 He

Q = BE of product – BE of reactant = [28 MeV – 2(2.2 MeV)] = [28 – 4.4] = 23.6 MeV

4 7 . K =

A 4Q

A

48 =

A 450

A

A = 100

4 8 . 1 1 00 1 1n p e

4 9 . N = N1 – N

2 1 2t t

0N [e e ]

5 0 . 90 days 3 half lives, left 1

8i.e. 12.5%

Disintegrated 100 – 12.5 = 87.5%

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5 1 . N = N0

e–t 0N

10= N

0e–t n(1) – n(10) = –t

n10t

n2

T

1/2 t =33 days

5 2 . 63% or nearly 2/3

5 3 . R = N = A

w

mN

M

EXERCISE –II

1 . (A) L = nh

2(B) E =

2

2

13.6Z

n

(C)c Z

v137 n

(D) K = U

2

2 . 2 20

1 1 1R

1 n

, n =4

(a) 0 1 2 0 1 2

hc hc hc 1 1 1

or 0 1 2 3 0 1 2 3

hc hc hc hc 1 1 1 1

3 .2

2

1 1 1 113.6 13.6Z

1 9 9 n

4 . |K|= 13.6 2 2

1 1

1 2

= 10.2 eV

|U|= 27.2 2 2

1

1 2

= 20.4eV

L = 2h h h

2 2 2

5 .2

3

keU

3r ;

2 2

4

U ke mvF

r rr

mv2 r3 = ke2; m2v2r2= 2 2

2

n h

4

2

mr

n (constant), Also

62

4

nV

m

2

3

6

keU

m3

n

, K.E. = 6

2

3

1 nmv

2 m

6

3

nU

m ,

6

3

nT.E. U K.E.

m

then total energy 6

3

n

m

6 . Kmax

= 10.4eV ; = 1.7 eVE = 10.4 + 1.7 = 12.1 eV n=3 to n=1

12.1 = 1242.eVÅ

7 . Stopping potential frequency 1

wave length

Saturation current rate of photoelectron

emission. Also, K.E.max

= h– , P= 2mKE

8 . (1) A

A

h

2mT (2) B

B

h

2mT

(3) TB = T

A – 1.5 eV (4)

B = 2

A

9 . Kmax

= 1242

eV 4.5eV200

Kmax

= 1.7eV at cathodeK

max = (1.7 + 2) eV at anode

If polarity is revered, no e– reach at collector.

1 0 . N = N0e–t 1 / 2T0

0

NN e

2

e

1 / 2

log 2T

T

mean =

1

1 1 . Assuming to be+10

01

172 172 17669 7169

176 18072 74

E D C

B A

1 2 . At t = 0 : N1 = N

0

At time t : N2= N

0e–t

Decayed in time t (N1 – N

2) = N

0(1–e–t)

Probability that a radioactive nuclei does not decay

in t=0 to t :

tt0

0 0

N N ee

N N

1 3 . Add equation 2 41 23 H He + p + n

(A) Q=[– [m ( 42 He )+m(p)+m(n)]+3m( 2

1 H )]× 931 MeV

(B) 1016W = Q

t

1 4 . A is balanced both in mass number & atomic no.

1 5 . h = h0 + eV

s ; V

s =

0(h h )

e

; eV

s=h–

Here =frequency of incident light and= properties of emitter

1 6 . E1 = 2 2

1 2

hc 1 1

n n

For second excited state to first excited state

E1 =

hc 1 1 hc 5

4 9 36

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For first excited state to ground excited state

2

hc 1 hc 3E 1

4 4

(A) 1

2

E 5

E 27 (B)

1 2 2

2 1 1

E hc 27

E hc 5

(C) P 1

1

2

P 5

P 27

1 7 . P.E. = –2(K.E.)T.E. = (P.E.) + (K.E.)T.E. = –2(K.E.) + (K.E.)T.E. =–(K.E.)–T.E. = K.E. K.E. = 3.4 eV = 6.6 × 10–10 m

1 8 . 122.4 = 2

2

13.6Z

1 Z=3; 91.8 =

1122.4 1

4

So an electron of KE 91.8eV can transfer itsenergy to this atom.

1 9 . Room temperature n=1Upon absorption excitations take place to manyhigher states which upon de-excitation emit all U.V.,infrared and visible light.

2 0 . Room temperature n=1 so lyman series

2 1 . 2

27.2eVU C

n

2

13.6eVk

n

2

13.6eVE C

n E = 13.6eV 2 2

1 1

n m

here in these questions C = + 27.2 eV

2 2 . We have 1

r

, = reduced mass

2 22e N

e N

mv mvKq

r r r ; mv

er

e =

nh

2

2 3 . K > 20.4 eV for inelastic collision

2 4 . decay : 2He4, so both Z & A decreases.

decay : +1

e0 ,so A will not change but Z will change (decreases)decay :

–1e0 ,

so A will not change but Z will change (increases)decay : no change in A & Z.

2 5 . V = 10

min

1240010

volt

V = 10

4

12

12400 10 12.4volt 10 18.75

66.366.3 10

kV

2 6 . min

1

V if V then (

min)

No. of collision per electron increase then intensityincreases

2 7 .

N

Z

BE/A

A

2 8 . = 0.173 , N = N0e–t

N = (N0 – N

0e–t) = Decayed amount

N = N0

11

e

0 0N N (1 0.37) N (0.63)

0

0 0

N N (0.63)100 63%

N N

T1/2

= elog 24year

2 9 . Pr =

r

r

n hc

, Pb =

b

b

n hc

if Pr = P

b Since

r >

b n

r > n

b

3 0 . (a) P = N hc

(b) i = n e

(c) %

n100%

N

3 1 . We have work is done by only electric field. Thus if

E v, v decreases, & thus momentum of electron

decreases & vice–versa, while in magnetic field v

remains constant.

3 2 . For electron db

= 150

100 50=1Å

3 3 . In photo electric effect only one to one Interaction.

3 4 . 0t

0N N e , 010 t

0N ' N e 09 t1e

e

t = 0

1

9

3 5 . eV1 =

1

hc

, 2

2

hceV

, eV3 =

0

3

h

if 2V2 = V

1 + V

3

2 1 2

2 1 1

harmonic progression

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3 6 . = 5eV – 2eV = 3eV

–VS =

6eV 3eV

e

V

S = –3V

3 7 . n

nhJ

2

n

n = 2(r

0n2)

So Jn

n

n = (2r

0)n

3 8 . En =

2

2

13.6(Z )

n

22

2n 2 2

13.6 13.6 ZE Z

4(2n) n

E2n

– En =

2 2

2 2

13.6 Z Z13.6

4 n n

E2n

– En=

2 2

2 2

13.6 Z Z13.6

4 n n

E2n

– En=

2 2

2 2

Z 13.6 Z13.6 (10.2)

4n n

(E2n

)(En) =

2 4

4

(13.6) Z

4 n

2

22n n

2 42n n

4

Z(10.2)

(E E ) n

(E )(E ) (13.6) Z

4 n

2

2n n

22n n

(E E ) n

(E )(E ) Z

3 9 . 10 = nC2 n = 5; then 5 orbits are involved upon

coming to second excited stateso nth excited state is 6th [2nd, 3rd, 4th, 5th, 6th]

4 0 . On coming from 4 1 energy is greater than,less than or equal to energy corresponding to2 4

4 1 .

2 220 0

2 2 4

0 0

V VV

r n r n r n

4 41 2

4 42 1

a n 2 16

a 81n 3

4 2 . T n3

3

i

f

T 1 n n 1

T 27 m m 3

4 3 . No of spectral lines

2n(n 1)10 n n 20 0

2

n=5; n=–4

E = 13.61

125

=13.056; = hc

E=95nm

4 4 . (a) E = 13.62

2 2

1 1 hc 1mv

21 5

(b) 0 = mv – h

4 5 . Excitation upto n=3 is required so that visible lengthis emitted upon de-excitation.

So required energy = 13.6 1

19

= 12.1eV

4 6 .

21 1 1

22 22

r n 1 n 1

r 4 n 2n

31 1

32 2

T n 1

T 8n

4 7 . nth excited state = n 12

n(n 1)C

2

= n

4 8 .2 2

3

nh n ZL ; r 0.53 ; f

2 Z n

(f r L) Z Constant for all orbits.

4 9 . K K & K K, are type of atom.

5 0 . Energy released = (BE)product

– (BE)reactant

5 1 . 1t

1A Ne , 2t

2A 2 Ne

1 2( t t )1

2

2Ae

A

1

2

2An t

A

2

1

AT n

2At

n2

5 2 . Radioactivity law is valid for large samples

5 3 . 1T

1 0N N e ; 2T

2 0N N e

R1 = N

1 ; R

2 = N

2

(N1 – N

2) = 1 2

1 2

(R R )N N

T = e elog 2 log 2

;T

(N1 – N

2) =

1 2

e

(R R )T

(log 2)

(N1 – N

2) (R

1 – R

2)T

5 4 . Final product state

5 5 . 1 1 2 2

dNN N

dt 1 2t t

1 10 2 20N e N e

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EXERCISE –III

Match the column

1 . 2

1 1 vv ,KE , J n,

n rn

but r n2 and v 1

n 3

1

n

2 . For given atomic number, energy and hencefrequency of K–series is more than L–series. Inone series also –line has more energy or frequencycompared to that of –line.

3 . Consider two equations

eVs =

1

2mv2

max = hv –

0...(i)

no. of photoelectrons ejected/sec Intensity

h...(ii)

(A) As frequency is increased keeping intensityconstant.

|Vs| will increases,

1

2m(v2

max) will increase.

(B) As frequency is increased and intensity isdecreased.

|Vs| will increase,

1

2m(v2

max) will increase and

saturation current will decrease.(C) Its work function is increased photo emission

may stop.(D) If intensity is increased and frequency is

decreased. Saturation current will increase.

4 . (A) In half life active sample reduce = 0

R

2

Decay number of nuclei is = 0R

2

(B) N = N0e–t

where = decay constant, = 1 / 2

n(2)

t

1/2

1/2

tn2

t n2 00

NN N e N

e

00 0

0

N Ne 1 1 eN N N

e N e

(C) 1 / 2

0t / T

NN

(2) N =

0 03 / 2

N N

2 2 2

Comprehens i on–1

1 . For Balmer series, n1 =2 (lower) ; n

2=3,4 (higher)

In transition (VI), Photon of Balmer series isabsorbed.

2 . In transition II : E2 = –3.4 eV, E

4 = –0.85 eV

E = 2.55 eV; E = hc

=

hc

E= 487 nm.

3 . Wavelength of radiation = 103 nm = 1030Å

E = 12400

1030Å12.0eV

So difference of energy should be 12.0 eV (approx)Hence n

1 =1 and n

2=3

(–13.6)eV (–1.51)eV Transition is V.

4 . For longest wavelength, energy difference shouldbe minimum.So in visible portion of hydrogen atom,minimum energy is in transition VI & IV.

Comprehens i on–2

1 . i= qf = 2e v ev

3 2 r 3 r

2 . Mu=

evr

3

3 . net u d d

evr evr evr 2evrM M M M

3 6 6 3

4 . Net magnetic moment in that case will be zero.

Comprehens i on–3

1 . Q = CV 0Ane V

d

12

3 19

2.85 10 10n 16

0.5 10 1.6 10

n = 8.85 × 109

2 . Equivalent resistance

6

V 16VR

I 2 10 A

= 8 × 106

3 .nhc

P

where n = number of photons incident

per unit time. Also I = ne Ihc

Pe

6 34 8

6 19

(2 10 )(6.6 10 )(3 10 )

(4 10 )(1.6 10 )

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= 79.9

101.6

m 9900

1.6 Å = 6187 Å

Which came in the range of orange light.

4 . Stopping potential VS = 8V and KE = eV

S

KE = 8eV

Comprehens i on–4

1 . Rate of production of B depends on the decaying

rate (A) : AN

A 1

dN N

dt

B is decaying simultaneously with two rates

BN

2 B 2 2

dN N

dt

BN

3 B 3 2

dN N

dt

Number of nuclei of 'B' is = 1N

1 –

2N

2 –

3N

2

2 . B will increase when N1 > (

2 +

3)N

2

as initially N1 = N

2 = N

0

1>

2 +

3

3 . N2 = 0, N

1=0 as both will decay completely :

N3 =

0 2

2 3

2N

N

therefore B is incorrect

EXERCISE –IV(A)

1 . Number of photons falling/s

2

1n

r [for point source]

So for new distance n' = n

9

ss

I 18mAI 2mA

9 9

Also saturated current nV

d is independent of n.

(i) VS = 0.6V (ii) I

s=

2

2

18 0.22mA

0.6

2 . E330

=12400

3.7575eV3300

E220

=12400

5.636eV2200

Let work function be eV

0 = (3.7575 – ) e, 2eV

0 (5.6363–)e

V0 = 1.87 V

3.(i) For metal 1 threshold wavelength is 1

1

1= 0.002 nm–1 or

1 = 500 nm = 5000Å

For metal 2 threshold wavelength is 2

1

2= 0.005 nm–1 or

2 = 200 nm = 2000Å

(ii) = hc

0or

1

0

1 :

2 = 2 : 5

(iii) Metal 1 because 1 lies in visible wavelength range.

4 . We have eV0, 0

hc hh V

e e

From graph, 2V 2eVe

Also slopes of the graph = 15

2 h

e0.49 10

1934

15

2 1.6 10h 6.536 10 J / sec

0.49 10

5 . Maximum kinetic energy of photo electrons

Kmax

= 1

2mv2

max=

hc

–

0

Now let 3000Å = then 6000Å = 2

20

max 1

2

max 20

hcv 9

hc 1v2

34 8

0 10 19

7hc 7 6.62 10 3 101.81 eV

16 16 3000 10 1.6 10

6 . KEmax

= h , Also Ephoto/time

= N hc

(i) When intensity of light is decreased numbr ofphotons decreases but KE

max remains same

(ii) When emitting surface is charged, changes. SoKE

max changes . If emitter is changed then +ve no.

of photoelectrons becomes zero.

(iii) Increasing , hc

7 . Energy incident / area of sphere / sec = 210

4 0.1

Energy incident on atom

= 29 19

2

100.05 10 6.25 10 J

4 0.1

Energy of one photon = hc

12.525eV

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No. of photon striking atom/sec

=

19

19

6.25 100.3125

12.525 1.6 10

No. of photons/ area

20

2 19

10 1 10 10 10

8012.525 1.6 104 0.1

Since n = 1% 2010

n80

8 . de–broglie

= h h

mv2mK

v= u + at = eE

0 tm

h

eEt 2

d h

dt eEt

9 . From the figure it is clear that

(P+1). /2 = 2.5 Å

/2 = (2.5 – 2.0)Å = 0.5 Å

or = 1 Å = 10–10 m.

de–Broglie wavelength is given by

h h

p 2Km

K = kinetic energy of electron

2 34 217

2 31 10 2

h (6.63 10 )K 2.415 10 J

2m 2(9.1 10 )(10 )

17

19

2.415 10eV

1.6 10

= 150.8 eV

10.(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 h

P 2Km

K = kinetic energy of electron

Substituting the values, we have

34

19 31

(6.6 10 J s)

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

= 6.63 × 10–10 m or = 6.63 Å

1 1 . (i) Magnetic moment ehn

IA4 m

Produced is nth orbit

for hydrogen n =1 Magnetic moment eh

4 m

(ii)eh

M B4 m

× B × sin30° =

ehB

8 m

1 2 .1 hc 3hc

E3 E

when E' = E, '= hc

E 3

So new wavelength = 3

1 3 . From the given conditions :

En–E

2=(10.2+17)eV=27.2eV

and En – E

3 = (4.25 +5.95) eV = 10.2 eV

Equation (1) – (2) gives

E3 – E

2 = 17.0 eV or Z2(13.6)

1 1

4 9

= 17.0

Z2(13.6) (5/36) = 17.0 Z2 = 9 Z = 3

From equation (1) Z2(13.6) 2

1 127.2

4 n

(3)2 (13.6) 2

1 127.2

4 n

2

1 1

4 n = 0.222

1/n2 = 0.0278 n2 = 36 n = 6

1 4 . 2B B

1 1 1 1 RR

42

p

1 1 1 7RR

9 16 144

; B p

7:

36

1 5 . Assuming Bohr's model to be applicable to the Heatom too;

Energy of electron = –13.6 × 4 eV= – 54.4 eV

initial energy of electron = 0

Energy of photon emitted = 544 eV

hc22.8mm

E

16. (i)Operating voltage = 40 kV, 0.5% energy for x ray

99.5

n e V 720100

14 3

720n

1.6 10 0.995 40 10

= 1.1 × 1017

(ii) Velocity of incident e– 1

2mv

e2 = eV

ve =

19 3

31

2eV 2 1.6 10 40 10

m 9.1 10

= 1.2 × 108 m/s

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1 7 . Given K

= 71 pm = 0·71 Å

EK – E

L =

hc

K

= 12400

0 7117 46

eV Å

ÅkeV

–.

Thus EL

= EK

– 17·46 keV = 23·32 keV – 17·46 keV = 5·86 keV

1 8 . Total mass annihilated e 2

12m MeV

c

Total energy produced=mc2=2

2

1MeV c

c =1MeV

Energy of 1 photon = 0.5 MeV

12hc 1.2 10

2.4pmE 0.5

1 9 . 2 2 41 1 2H H He Q

Q = [ 4 × 7 – 2 × 2 × 1.1] MeV = 23.6 MeV

2 0 . min

=hc

eV=

12400

VÅ

At 40 kV : min

=12400

40000= 0.31 Å

Wavelength of K is independent of applied

potential.

For K X–ray : 2

K

3 hc13.6 Z 1 E

4

K =

1216

12

Z b gÅ and given that

K =3

min

1216

12

Z b g = 3 × 0.31

2 1216Z 1 1308 Z 1 36 Z 37

0.93

2 1 . N1 = N

01e–t and N

2 = N

02e–3 t

1

2

N

N

N e

N e

t

t01

023

= e2t N01 =N

02 , for

N

N1

2=

e

1]

1 = 2t t=

1

2

2 2 . Activity of x = Activity of y or nx

x = n

y

y

nx

0 693

1 2

.

( )/T x

LNM

OQP = n

y

0 693

1 2

.

( )/T y

LNMM

OQPP

n

nx

y =

( )

( )

T

T

x

y

1 2

1 2

= 3

27 =

1

9

2 3 . After t time active fraction

= 1 – 0·36 = 0·64 = N

N0 = e–t

Now at t/2 time, active fraction= (e–t)1/2 = (0.64)½ = 0.8So decayed fraction in t/2 time is 0·2 or 20%

2 4 . 238 234 492 90 2U Th He

m= (238.05079 – 4.00260 –234.0 4363) uE = mc2 = 4.24764 MeVIf it emits proton spontaneously, the equation is notbalanced in terms of atoms & mass number.

238 237 492 91 1U Pa H

m =(238.05079–237.065121–1.007834)u= –0.022165 um is negative, so reaction is not spontaneous.

2 5 . Let at t = 0, capacitor starts discharging then at timet, activity of radioactive sample = R = R

0e–t

charge on capacitor = Q = Q0

e–t/RC

Now R

Q =

R e

Q e

t

t RC0

0

–

–

= R

Q0

0

et

Rc– – 1e j

It is independent of time if = 1

RC

R = 1

C =

t

C

avg =

20 10

100 10

–3

–6 = 200

EXERCISE –IV(B)

1 .

1

=hf/2

2

=hf/3

Potential at which electron stop coming out

from sphere–1, V1=

hfhf

hf2

e 2e

from sphere–2, V2=

hfhf

2hf3e 3e

After connection

(i) V + V = V1 + V

2 (V= final common potential)

2V = hf 2hf

2e 3e

7hfV

12e

(ii) For sphere–2 : k Q 2 hf 7hf hf

R 3 e 12e 12e

No. of electrons flows n = 2

Q hfR

e 12ke

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2 . Power of source = 2J/sec

Energy of 1 photon = hc 12400

2.067eV6000

No. of photons emitted /sec

=

19182 10

6.0474 102.067 16

No. of photons striking on sphere of 0.6 m

=

182

2

6.0474 10/ m

4 0.6

No. of photons passing through aperture

=

218 6

2

6.0474 10 0.1 4.2 10

2 44 0.6

photons

No. of photons incident / area on screen (assumingaperture to act as a secondary point source)

16

2

4.2 10

4 4 5.4

= photon flux

= 1.1462 × 1014 photons / m2 secNo. of photons incident on detector /sec

= 1.1462 × 1014 × 0.50

10000

Photo current =10N e 2.063 10 A

When the lens of f = –0.6 m is used,

1 1 1 1 1v 0.3m

v u f v 0.3

Thus image will be at 0.3 m from the lens in thedirection opposite to the screen.Distance between screen & image = 5.7 mNo. of photons striking lens = No. of photons striking the aperture = 1.05 × 1016 photons.Photons transmitted through the lens = 0.8 × 1.05 × 1016=0.84 × 1016 photon/sThis new situation, A point source emitting0.84 × 1016 photons/ sec of = 6000Åis kept at 5.7 m away from the screen.Thus photons striking / area of screen

=

16

2

0.84 10

4 5.7

= photon/sec =2.0574 × 1013

Electrons emitted=0.9 × 2.0574 × 1013× 0.5

10000

Current = 1.4813 × 10–10 AE

p = 2.067 eV ; =1eV

K.E.max

= 1.067 eV; VS = 1.067 V

3 . Energy of photo with 1

1

hc3000Å 4.14eV

Energy of photo with 1

2

hc1650Å 7.53eV

Power of source = 5 × 10–3W

315

2 19

5 10N 4.15 10

7.53 10 1.6

181 -19

1N for 1W = 1.5 10

4.14 1.6 10

Current = 4.8 × 10–3A

316

19

4.8 103 10

1.6 10

162

18

3 102 10 2%

1.5 10

15 132 4.15 10 0.02 8.3 10

Current n e 13.3 A

Also vmax 1650

= 2Vmax 5000

max 1650 max 3000

KE 4KE

4( 4.14 – ) = (7.53–) 16.56 – 7.53 = 3

= 9.03

3.01eV3

th

hc4126Å

E

4 .1

hc 12400eV 3.1eV

4000

2

hc 12400eV 2.48eV

5000

3

hc 12400eV 2.06eV

6000

Light having wavelength 6000 Å will not be ableto eject electrons.

Photo–current = 1 2(n n )e

where 1n

= no. of photons of light incident havingwavelength 4000Å

2n = no. of photons of light incident having

wavelength 5000Å. Here

330

1 2 3

1 2 3

hc hc hc I 3 10n n n 10

3 3

Area intercepted = 2cm2 = 2 × 10–4m2

Photo–current = 1 2(n n )e

3 4 10 19

34 8

(10 2 10 )(5000 4000) 10 1.6 10

6.63 10 3 10

= 0.144 A

5 . Momentum of particles = 1 2

h h&

Since particles are identical, mass of both particlesare same.

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mv1=

1

h

& mv2 =

2

h

; 1 2

COM

v vˆ ˆv i j2 2

2 2

1 21 21COM 1COM 2COM

v vv vˆ ˆv i j v v2 2 2

2 2 2 21 2 1 2

h 2h 2h

mv m v v mv mv

1 2

2 2 2 2

1 2

1 2

22h

h h

6 . (i) E = 47.2 = 13.6 Z21 1

4 9

Z = 5

(ii) E = 13.6 × 251 1

9 16

13.6 25 7

eV9 16

= 16.5 eV

(iii) E1

= 13.6 Z2eV=hc

12400eV A

13.6 25eV

= 36.4Å

(iv) (KE)1st orbit

= 1

2mv2

31 6 2

19

9.1 10 (2.2 10 5)

2 1.6 10

= 344 eV

(PE)1st orbit

= –2KE = –688eV

L = nh h

2 2

(n=1)

(v) Radius = (R0)

2 2n 0.53 1

z 5

= 1.06×10–11m

7 . Let n = no. of level of excited state nC

2 = 6 (spectral lines)

n = 3 (3rd excited states) No. of excited state level = n+1 = 4

2.7eVn=4

n=3

n=2

n=1

(Level B)

(Level A)

(i) Principal quantum no. of initially excited level B =4

(ii) 2.7 = k1 1

4 16

16k 2.7 14.4eV

3

Ionisation energy

= k1

1

= k = 14.4eV= 23.04 × 10–19J

(iii) Emax

= 4 1 = 13.5 eVE

min = 4 3 = 0.7 eV

8.(i)

2 2

20

mv 1 e

r 4 r

mvr=h

2 E

total = –

2

0

e

8 r

(ii)dE

Pdt

(loss of energy per sec)

204

0

d e P

dt 8 r r

20

2 40

e dr P

dt8 r r

0

r t

2 20 0

r 0

e r dr 8 P dt

r3 = r0

3 0 02

6P (4 )t

e

r =r

0

1 / 32e

30

3cr t1

r

(iii) For r=0, (to collapse and fall into nucleus)

1 –

2e

30

3cr t0

r

t=3 30 100

2 8 30e

r 10 10 100sec

813cr 3 3 10 9 10

9.(i)

2 2

n 2n 0 n

nh mv 4emvr and

2 r 4 r

2 2 22

0 0n 2 2

h n hnr

m4 e 400 me

(iii) th

2

2n

Z mE 13.6

n

2

2 2

1 1E 13.6 4 100 408eV

2 4

1 0 . Let the age of Earth be tyear

and initially both werepresent as N

0;

Nu238

= 31 / 2

0 0

t / t t / 4.5 10

N N

2 2

8

0235 t / 4.3 10

NNu

2

9t 1.18 10238

235

Nu 1402

Nu 1

t= 6.0418 × 109 years

1 1 . X A B

No. of neucleus disintegrated = txxt 1 e

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For 1

t

Disintegration = 1x x1 1 e

e

Energy released = 0

E x

e

Energy utilized for melting = 0

E x0.5

e

Mass of ice melted = 0

F

E x

2e L

1 2 . 2 21 D 3

1 T + 11 P

Mass defect M =MProduct

–MReactant

3.016049 1.00785 2 2.014102

= 4.023899–4.028204m = 4.3 × 10–3 amu and 1 amu 931.5E = mc2 = 4.01 MeV

3 21 1T D 4 1

2 0He n

m(mass defect) = Mproduct

– MReactant

= [4.002603+1.008665] – [3.016049+2.014102] = [5.011268 – 5.030151]

m = 0.018883E = m(931.5) 17.58 MeV

Edeutron

=total

E7.2MeV

3

Tota l Energy

To ta l M ass1 2

21

m mn

3 D

0.004305 0.018883n 100

3(2.014102)

= n = 0.384%

1 3 . (Take a sample of 1020 Cm atoms)

–decay 96

Cm248

96Pu244 +

2He4

Here m= (248. 072220 –244.064100 –4.002603) u = 0.005517 u

E=mc2 = 0.005517 × 931 = 5.136 MeV

Energy released in the decay of one atom

E = Efission

+ E = 0.08 × 200 + 0.92 × 5.136

= 20.725 MeV

Total energy released from the decay of all 1020 atoms

= 20.725 × 1020 MeV = 3.316 × 108J

Power output = Total energy released

mean life

= 8

13

3.316 10

10

= 3.3 × 10–5 watts

1 4 . Let the amount of 222xR be N

0

N7.6

= 0 0

2

N N

42

Nuclei left after 7.6 day = 0

N

4Decay constant

= 1 / 2

n2 0.693

t 3.8 24 3600

/sec

= 2.11 × 10–6 sec

0N

Activity No. of particles4

No. of neutrons produced

= 6

60N 2.11 101.2 10

4 4000

N

0= 9.095 × 1015 nuclei= 3.354 × 10–6g

1 5 . N1= M

0e–t

Thus momentum of this mass= Mv, v= velcity of mass at time t = M

0e–tv

Let at time dt, defect mass is ejected with relativevelocity v

0

veject/gram

= – v0 + v

Linear momentum of M at time (t+dt) = (M–dm) (v+dv) + dm(v–v

0)

Since fext

=0(M–dm) (v+dv) + dm (v–v

0)= Mv (M–dm) dv = dmv

0

0 0 0v dm v r vdv dm

dvM dm dt dt M dm M rdt

00

0

Mv v n

M rt

But M0–rt = M

0e–t;

00 0t

0

Mv v n v t

M e

1 6 . Production of radioactive nuclei = /secDisintegration= N

A

At any time t =A

A

dNN

dt

AN t

tAA

A0 0

dNdt N 1 e

N

Nuclei disintegrated = tA

1t N t 1 e

Total energy produced = t0E t 1 e

Energy used in water heating

t00.2E t 1 e ms T

t00.2E t / 1 e

TmS

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EXERCISE –V(A)

ATOMIC STRUCTURE & X-RAY1 . Energy required to ionize a atom from nth orbit is

= + 2

13.6eV

n

2 2

13.6E eV 3.4eV

2

2 . Energy required to remove an electron from an orbit

is + 22

13.6 Z

n eV . So, to remove the electron from

the first excited state of Li2+ is

2

2

13.6 3E

2

=3.4 × 9 = + 30.6 eV

3 . Ionization potential will be lowest for the atom inwhich the electrons are the farthest from thenucleus. So, the atom with the largest size will havethe electron the farthest from the nuclei, hence toremove the electron from this atom will be easiest.

So, the atom with least ionization potential is 13355 Cs .

4 . The wavelengths involved in the spectrum of

deuterium 21 D are slightly different from that of

hydrogen spectrum; because masses of two nucleiare different.

6 . The energy of emitted photon is directly proportionalto the difference of the two energy levels.Thisdifferences is maximum between level (2) and level(1) hence photon for maximum energy will beliberated for this transition only.

7 . Whenever an atom gets de-excited from higher tothe lower orbit, it emits radiation of a given frequency

E hf f E

The photons of highest frequency will be emitted,for the transition in which E is maximum, i.e., from2 to 1.

8 .2

2k mvmv k

r r (independent or r)

hn mvr r n

2

and 21

T mv2

is

independent of n.

1 0 . Rquired energy = 13.6 (Z)2 2 2

1 2

1 1

n n

eV

= 13.6 (3)2 2 2

1 1

1 3

= 108.8 eV

1 1 . Number of lines = nC2 = 4C2 = (4)(3)

(2) = 6

1 2 . Energy =

2 2

2

L (n )

2I 2( r )

where µ = reduced mass = 1 2

1 2

m m

m m

so energy = 2 2

1 2

2

1 2

n (m m )

2m m r

1 3 . Energy of radiation emitted

E = h 2 2

0 02 2 2 2

1 1 2n 1E Z E Z

(n 1) n n (n 1)

20 4 3

2n 1h E Z

n n

PHOTOELECTRIC EFFECT

1 4 . Work function Na Cu

0 0

0 Cu Na

WhcW hf

W

Na Cu

Cu Na

W 4.52

W 2.3

1 5 . Covalent bonding electron cloud overlappingregion electron probability density wave nature of electron

1 6 . 21 0 1

1hf mv

2 , 2

2 0 21

hf mv2

2 21 2 1 2

2hv v f f

m

1 7 . Total change in momentum from a reflect ingsurface = 2P

P

P

The momentum of incident radiation 2E

Pc

Hence, momentum transferred =2E

c

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1 8 . According to Einstein's equation of photoelectriceffect hf=W0+Kmax

Kmax = (h)f–W0

q

Kmax

tW0

The slope of K vs f graph is h which is a fundamentalconstant and same for all metals at all intensifies.

1 9 . W0=4eV, max = ? ; 0

max

hcW

max

12404eV nm eV

hc= 1240 nm-eV

max = 310 nm.

2 0 . Photocurrent Intensity

Intensity due to point source at r m is 2

1I

r

As distance is reduced to half, the photocurrent dueto point source will increase by a factor of 4.

2 1 . The de-Broglie wavelengths and kinetic energy ofa free electron are related as

2 1 12

1 2

Kh 1

K 22mK 2

2 2 . According to Einstein's theory of photoelectric effectEnergy of incident photon = W0 + eV0

W0 = 6.2 eV; eV0 = 5 eVEnergy of incident photon = 11.2 eVThe wavelength corresponding to this energy is 110nm which falls in the ultraviolet region.

2 3 . The photoelectric phenomenon is an instantaneousphenomenon, hence the time taken by an electronto come out of metal is approximately 10–10 sec(found experimentally).

2 4 . With the increase in wavelength energy of thephoton decreases. Therefore the KE of the electroncoming out from the cathode also decreases. Dueto which there will be a small decrease in platecurrent and once becomes more than thresholdwavelength electron will not come out from thecathode and hence current will become zero.

2 5 . The relation between energy (E) of a photon andmomentum (P) associated with the photon is

E = pc

The corresponding momentum E hv

pc c

2 7 . 2d cos i = nh

2d cos imeV

V = 50 volt

2 8 . 2d cos i = ndB

2 9 . k 0 0

hc 12400E 1.68

4000

By solving it 0 = 1.42 eV

3 1 . No. of photons emitting per second from a sourceof power P is n = (5 × 1024) P

wavelength emitting 24

n nhcor

P5 10 P

209

24 3

100.5 10 m 50Å

5 10 4 10

And this wavlength comes in X ray region.

Radioactivi ty Nuclear Physics

3 5 .

n

0

N 1

N 2

where n are is number of half-lives.

If T1/2 = 5 years then in 15 years, n=3

3

0

0

NN 1 1N

N 2 8 8

3 6 . Though the compound can emit all the four particlesnamely electrons, protons, He2+ and neutrons. Butthe particle neutron can't be deflected in themagnetic field, since it is a neutral particle. Hence,the deflectable particles are protons, electrons andHe2+.

3 8 . 238 234 42U U He

Let recoil speed be V then by COLM

4u234V 4u V

234

3 9 . Rate of disintegration at any instant is directlyproportional to the number of undecayed nuclei at

that instant, i.e., dNN

dt

Where = decay constantWe are given that at t=0

0

dN disintegrationN 5000

dt minute

at t =5 minute

dN disintegration- N 1250

dt minute

t0N N e

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On multiplying both sides by we get

(N) = (N0)e–t 1250 = 5000 e–t

t1e

4

Taking logarithm on both sides we get

ln1–ln4=e–(5) 0–ln4=–5 21

ln 25

= 0.4 ln2.

4 0 . 8 have been emitted4– have been emitted2+ have been emitted reduces atomic number by 2– increases atomic number by 1+ decreases atomic number by 1So, Zeff = 92 – (8×2) + (4×1)–(2×1)= 96–18= 78

4 1 . A nucleus during decay emits (He2+), (electrons), or neutrino; it does not emit protons during decay.

4 2 . In order to fuse two nuclei against repulsion, therepulsive potential energy has to be supplied bykinetic energy i.e., PE=KE

7.7 × 10–14 =3

2kT

7.7 × 10–14 =3

2× 1.38 × 10–23T

T = 3.7 × 109 k

4 3 . Applying conservation of momentum, we get

1 1 2 2m v m v

1 2

2 1

m v 1

m v 2

Also, 1/3 3R m m R

3 1/ 3

1/ 31 1 1

2 2 2

m R R1 11 : 2

m 2 R R 2

4 4 . Given that 2 2 41 1 2H H He energy

Total B.E. of deuterium nucleus = 2.2 MeVTotal B.E. of He nucleus = 28 MeVOn conservating energy on both sides we get(Energy)Deutron × 2 = (Energy)He+ Energy released4.4 = 28 + EE = 28 – 4.4 = 23.6 MeV

4 5 . At the closest point of approachInitial KE = Final PE

5 × 106 × 1.6 × 10–19 =1 2

0

kq q

r

9 19 19

0 6 19

9 10 92 1.6 10 2 1.6 10r

5 10 1.6 10

= 5.3 × 10–14 m = 5.3 × 10–12 cm.

4 6 . Intensity of gamma radiation when it passes through

x is x0I I e

360

0

II e

8

...(i) and

x00

II e

2

...(ii)

36 x 3 x 361 1e and e e e x 12 mm

8 2

4 7 . We know that,

n

0

N 1

N 2

where, N are the

radioactive nuclei left after n-half livesN0 are the initial number of nuclei and n is number

of half-lives,

3 n

0

0

N / 8 1 1 1n 3

N 8 2 2

Therefore T1/2 = 15

3=5 min.

4 8 . Radius of a nuclei (Mass Number)1/3

1/3

Te

Al

R 125 5

R 27 3

Te Al

5R R 6

3 fermi

4 9 . 73X n, Li

1 A 7 40 Z 3 2n X Li He

On conservating atomic number and mass numberson both sides, we getA = (7+4) –1 = 10Z = (3+2) – 0 = 5

Hence, A 10Z 5X B

5 0 . Charge on -particle = 2eCharge on target nucleus = ZeWhen the -particle is projected towards the targetnucleus, then at r=r0, the -particle comes tomomentary rest. This position r0 from target nucleusis known as distance of closest approach.Applying law of conservation of energy, we get

2

0

K ze 2e1mv

2 r 0 0 02

1 1r , r Ze, r

mv

5 1 . 7 1 83 1 4Li P Be

The particle that will be emitted with Beryllium willbe Gamma radiations.

5 2 . The energy spectrum of -particles emitted from aradioactive source is

E

N(E)

E0

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5 3 . In a nuclear reaction the energy remains conserved

7 43 2p Li 2 He

energy of protons + 7(5.60) =2(4×7.06)

energy of proton = 17.28 MeV

5 4 . Nuclear binding energy =

[Expected mass of nucleus-Actual mass of nucleus]c2

Expected mass of nucleus = 8Mp + 9Mn

Actual mass of nucleus = M0

Nuclear binding energy = [8Mp+9Mn–M0]c2

5 5 . Gamma ray is an electromagnetic radiation, due to

the emission of gamma ray, neither the mass number

not the atomic number changes. Though the

daughter nucleus is same as parent nucleus but still

there is a difference in the two, i.e., the daughter

nucleus so obtained is present in one of the excited

states and not in the ground state.

5 6 . Given that

1/2 av YxT T

X Y

0.693 1

Y X

1

0.693

Y X Y X1.44 R 1.44 R

As decay rate of Y is more than the element X;

hence Y will decay faster than X.

5 9 . Total kinetic energy of products

= Total energy released 2 2p p

2m 2m

= (mass defect) c2 (where m =M

2given)

22p M M

2 (M m) c2m 2 2

2

2p2 m c

M2

2

2

2

M2 v

2 m2 m c v cM M

6 0 . Because energy is releasing Binding energy per

nucleon of product > that of parent E2 > E1.

6 1 . ZXA

3( )24 Z–6( )

A–12 2( )+10 Z–8( )A–12

A 12 Z 8No. of neutrons A Z 4

No. of protons Z 8 Z 8

6 2 .

t / T

0

N 1

N 2

2t

T1 1

3 2

&

1t

T2 1

3 2

2 1

1( t t )

T2 1( t t )1 1

12 2 T

T = t2 – t1

t2 – t1 = 20 min.

6 3 . M+ mN m5m'p'

p

According to Conservation of linear momentum

P' = P. Therefore ' =

6 5 . Released energy

= [1.6747 × 10–27 – 1.6725 × 10–27 – 9 × 10–31]

× (3 × 108)2 J = 0.73 MeV

EXERCISE –V-B

1 . The continuous X–rayspectrum is shown in figure.All wavelengths >

min

are found, where

E

min

min

12375Å

V(in volt)

Here, V is the applied voltage.

2 . (in Å) = 12375 12375

Å 3093 ÅW (eV ) 4.0

=

309.3 nm = 310 nm

3 . Number of nuclei decreases experientially

N = N0e–t

and Rate of decay

dN

dt

=N

Therefore, decay process lasts upto t=. Thereforea given nucleus may decay at any time after t=0.

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4 . Since, the wavelength () is increasing we can saythat the galaxy is receding. Doppler effect can be

given by–

c v'

c v

706 = 656

c v

c v

...(i)

2c v 706

1.16c v 656

c + v = 1.16 c – 1.16 v

0.16c

v2.16

80.16 3.0 10

m / s2.16

v 2.2 × 107 m/s

If we take the approximation then equation (i) can be

written as = v

c

...(ii)

From here v .c

8

706 656(3 10 )

656

m/s

v = 0.23 × 108 m/sWhich is almost equal to the previous answer.

So, we may use equation (ii) also.

5 . Atomic number of neon is 10.

By the emission of two – particles, atomic numberwill be reduced by 4. Therefore, atomic number of

the unknown element will be Z = 10 – 4 = 6Similarly mass number of the unknown element willbe A = 22 – 2 × 4 = 14

Unknown nucleus is carbon (A = 14, Z=6).

6 . From law of conservation of momentum,

P1 = P

2 (in opposite directions)

Now de–Broglie wavelength is given by h

p

[h = Plank's constant]Since, momentum (p) of both the particles is equal,

therefore 1 =

2

1/

2 = 1

7 . Both the beta rays and the cathode rays are made up

of electrons. So, only option (a) is correct.(b) Gamma rays are electromagnetic waves.(c) Alpha particles are doubly ionized helium atoms and

(d) Protons and neutrons have approximately thesame mass.

8 . (t1/2

)x = (t

mean)y

x y

0.693 1

x = 0.693

y

x <

y Rate of decay = N

Initially number of atoms (N) of both are equal butsince

y >

x, therefore, y will decay at a faster rate

than x.

9 . Radius of a nucleus is given byR = R

0A1/3 (where R

0 = 1.25 × 10–15 m)

= 1.25 A1/3 × 10–15mHere, A is the mass number and mass of theuranium nucleus will be m Am

p

mp = mass of proton = A(1.67 × 10–27 kg)

Density = 27

15 33

mass m A(1.67 10 kg)

4volume A(1.25 10 m)R

3

2.0 × 1017 kg/m3

1 0 . Energy is released in a process when total bindingenergy of the nucleus (= binding energy per nucleon× number of nucleons) is increased or we can say,when total binding energy of products is more thanthe reactants. By calculation we can see that only incase of option (c), this happens. Given : W 2YBinding energy of reactants = 120 × 7.5 = 900MeVand binding energy of products= 2(60 × 8.5) = 1020 MeV > 900 MeV.

1 1 . In hydrogen atom En = 2

Rhc

n . Also E

n m

where m is the mass of the electron.Here, the electron has been replaced by a particlewhose mass is double of an electron. Therefore, forthis hypothetical atom energy in nth orbit will be

given by En = 2

2Rhc

n

The longest wavelength max

(or minimum energy)photon will correspond to the transition of particlefrom n=3 to n=2.

3 2 2 2max

hc 1 1E E Rhc

2 3

max

= 18

5R

1 2 . n

1

n KE 2

1

n (with positive sign)

Potential energy U is negative and

Un

n

1

r

2

n0 n

1 ZeU .

4 r

2

1

n

[because rn n2] (with negative sign)

Similarly total energy En 2

1

n(with negative sign)

Therefore, when an electron jumps from someexcited state to the ground state, value of n willdecrease. Therefore, kinetic energy will increase(with positive sign), potential energy and total energywill also increase but with negative sign. Thus, finallykinetic energy will increase, while potential and totalenergies will decrease.

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1 3 . Minimum wavelength of continuous X–ray spectrum

is given by min

(in Å) = 12375

E(in eV )

Here, E = energy of incident electrons (in eV)

= energy corresponding to minimum wavelength

min

of X–rays E = 80 keV = 80 × 103 eV

min

(in Å) = 3

12375

80 10 0.155

Also the energy of the incident electrons (80 keV) is

more than the ionization energy of the K–shell

electrons (i.e. 72.5 keV). Therefore, characteristic

X–ray spectrum will also be obtained because

energy of incident electron is high enough to knock

out the electron from K or L–shells.

1 4 .1

2

x

x

N (t ) 1

N (t) e

10 t0

t0

N e 1

eN e

(Initially both have same number of nuclei say N0)

e =–t/e–10t e = e9t, x1

= 10 and x2

=

9 t=1 t = 1

9

1 5 . Energy of infrared radiation is less than the energy of

ultraviolet radiation. In options (a), (b) and (c), energy

released will be more, while in option (d) only, energy

released will be less.

1 6 . Wavelength k is independent of the accelerating

voltage (V), while the minimum wavelength c is

inversely propor t ional to V. Therefore, as V is

increased, k remains unchanged where as

c

decreases or k –

c will increase.

1 7 . During –decay, a neutron is transformed into a

proton and an electron. This is why atomic number

(Z = number of protons) increases by one and mass

number (A = number of protons + neutrons) remains

unchanged during beta decay.

1 8 . The total number of atoms can neither remain

constant (as in option a) nor can ever increase (as in

options b and c). They will continuously decrease

with time. Therefore, (d) is the appropriate option.

1 9 . In second excited state n=3,

So, H =

Li =

3h

2

while E Z2 and ZH = 1, Z

Li = 3

So, |ELi| = 9 |E

H| or |E

H| < |E

Li|

2 0 .q ne

it t

i t

ne

Substituting i = 3.2 × 10–3A

e = 1.6 × 10–19 C and t =1s

we get, n =2 × 1016

2 1 .

n

0R R2

...(i)

Here R = activity of radioactive substance after n

half–lives 0R

6 (given)

Substituting in equation (i), we get n =4

t = (n)t1/2

= (4) (100 s) = 400 s

2 2 . During–decay atomic number (Z) and mass number

(A) does not change. So, the correct option is (c)

because in all other options either Z,A or both is/are

changing.

2 3 . U = eV = eV0 n

0

r

r

0dU eV

| F |dr r

this force will provide the necessary centripetal

force.

Hence,2

0mv eV

r r

0eVv

m ...(i)

Moreover,nh

mvr2

....(ii)

Dividing equation (ii) by (i).

We have : mr = 0

nh m

2 eV

rn n

2 4 . (rm) =

2m

z

(0.53 Å) = (n × 0.53)Å 2m

z=n

m =5 for 100

Fm257 (the outermost shell) and z=100

2(5) 1

n100 4

2 5 . Nuclear density is constant hence, mass volume

m V.

2 6 . Given that K1 + K

2 = 5.5 MeV ...(i)

From conservation of linear momentum, p1 = p

2

1 22K (216m) 2K (4m) as P 2Km

K2 = 54 K

1 ...(ii)

Solving equation (i) and (ii).

We get K2 = KE of –particle = 5.4 MeV

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2 7 . Saturation current is proportional to intensity whilestopping potential increases with increase in

frequency. Hence, fa = f

b while I

a < I

b

2 8 .1

2

h

2mEhc

E

or 1 1 / 2

2

E

2 9 . Activity reduces from 6000 dps to 3000 dps in140 days. It implies that half–li fe of theradioactive sample is 140 days. In 280 days (or

two half–lives) activity will remain 1

4th of the

initial activity. Hence, the initial activity of thesample is

4 × 6000 dps = 24000dps3 0 . The first photon will excite the hydrogen atom (in

ground state) in first excited state

(as E2–E

1 = 10.2eV). Hence during de–excitation a

photon of 10.2eV will be released. The secondphoton of energy 15eV can ionise the atom. Hence

the balance energy i.e.,(15 – 13.6)eV = 1.4 eV isretained by the electron. Therefore, by the secondphoton an electron of energy 1.4 eV will be released.

31.

2 21 2 22

2 1

1 Z 1 1 Z 1(Z 1)

Z 1 4 11 1

Solving this, we get Z2 = 6

3 2 . 4(2He4) =

8O16

Mass defectm = {4(4.0026) – 15.9994} = 0.011 amu Energy released per oxygen nuclei

= (0.011) (931. 48) MeV = 10.24 MeV3 3 . After two half lives 1/4th fraction of nuclei will remain

undecayed. 3/4th fraction will decay. Hence, the

probability that a nucleus decays in two half lives is 3/4.3 4 . The series in U–V region is Lymen series. Longest

wavelength corresponds to minimum energy which

occurs in transition from n=2 to n=1.

2 2

1

R1221 1

1 2

...(i)

The smallest wavelength in the infrared regioncorresponds to maximum energy of Paschen series.

2

1

R1 1

3

...(ii)

Solving equation (i) and (ii), we get = 823.5 nm

3 5 . Momentum of striking electrons h

p

Kinetic energy of striking electrons

K =

2

2

p h

2m 2m

This also, maximum energy of X–ray photons.

Therefore,

2

20

hc h

2m

2

0

2m c

h

3 6 . Rest mass of parent nucleus should be greater thanthe rest mass of daughter nuclei.

3 7 . As for continuous X–rays min

= hc

eV so cut off

wavelength depends on the accelerating potentialand is independent of nature of target.

3 8 . Activity = N & T1/2

=

n2

So 5 =

0

1

n2(2N )

T & 10 =

0

2

n2(N )

T T1 = 4T

2

3 9 . 550 nm light cannot emit electron for metal of work

function 3eV , so saturate current decreases for

P to Q to R. Also|Vsp

|>|Vsq

|>|Vsr|

4 0 . Energy incident = Power(time)

= (30 mW) (100 ns)

= 3 × 10–9

So momentum =

917

8

E 3 1010 kg m / sec

c 3 10

.

M C Q

1 . Due to mass defect (which is finally responsible forthe binding energy of the nucleus), mass of a nucleusis always less than the sum of masses of its

constituent particles. 201.0 Ne is made up of 10

protons plus 10 neutrons.

Therefore, mass of 2010 Ne nucleus,

M1 < 10 (m

p + m

n)

Also, heavier the nucleus, more is the mass defect.

Thus, 20 (mn + m

p) – M

2 > 10 (m

p + m

n) –M

1

10 (mp + m

n)>M

2–M

1M

2< M

1 + 10(m

p + m

n)

Now since M1 < 10 (m

p + m

n) M

2 < 2M

1

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2 . Time period,

Tn =

n

n

2 r

v

(in nth state) i.e. T

n

n

n

r

But rn n2 and

n

1

n

Therefore, Tn n3 Given T

n1 = 8T

n2

Hence, n1 = 2n

2

3 . From the relation,

eV = hc

hc 1

Ve e

This is equation of straight line.

Slope is tan = hc

e

1:

2:

3 =

01 02 03

hc hc hc: :

= 1 : 2 : 4

1

01

10.001nm

01 = 10000 Å

1

02

10.002nm

02 = 5000Å

1

03

10.004nm

03 = 2500 Å

Violet colour has wavelength 4000 Å.

So violet colour can eject photoelectrons frommetal–1 and metal–2.

4 . (A ) For 1 < A < 50, on fusion mass number forcompound nucleus is less than 100

Binding energy per nucleon remains sameNo energy is released

( B ) For 51 < A < 100, on fusion mass numberfor compound nucleus is between 100 & 200

Binding energy per nucleon increases

Energy is released.

(C ) For 100 < A < 200, on fission, the massnumber of product nuclei will be between50 & 100

Binding energy per nucleon decreases

No energy is released

( D ) For 200 < A < 260, on fission, the massnumber of product nuclei will be between100 & 130

Binding energy per nucleon increases

Energy is released.

5 Angular momentum = nh 3h

2 2

n = 3

Also r =

2

0

na

2

20

0

9a 3a Z 2

2 Z

For de-exci tation

2

2 2 2 2

1 2 1 2

1 1 1 1 1Rz 4R

n n n n

For n = 3 to n = 1:

1 1 14R

1 9

9

32R

For n = 3 to n = 2:

1 1 1 94R

4 9 5R

For n = 2 to n = 1:

1 1 1 34R

1 4 R

Match the column2 . (i) Energy of capacitor is less 1/2 cv2 therefore p.

(ii) work is done on the gas hence energy increasestherefore q.(iii) When mass decreases its energy increases.(iv) when current flows energy of magnetic field isgenerated therefore t.(B) work is done on the gas(C) mass is reduced and mass defect is convertedinto energy.(D) mass decreases due to mass defect.

Matching List Type1 . Completing reaction in list II

(1) 15 158 7O N + decay)

(2) 238 23492 90U Th (Alpha decay)

(3) 185 18483 82Bi Pb (Proton emission)

(4) 239 14094 57Pu La (Fission)

Comprehension Based Question

Comprehens i on#1

1 .

H

–3.4eV

–13.6eV10.2eV

He+

–3.4eV

–6.0eV10.2eV

–13.6eV

–54.4eV

For H atom E = 10.2 eV, This goes to exciteHe+ ion from n = 2 to n = 4

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2 . For visible region E < 10.2 eV and in this case

E = –3.4 – (–6) = 2.6 eV

=

hc

E=

12400eV Å4800Å

2.6eV 4.8× 10–7 m

3 . Ratio of kinetic energy

2

1 112

2 2 2

Z nK

K Z n

Since n1 = n

2 = 2 & Z

1 = 1

for H, Z2 = 2 for He+ 1

2

K 1

K 4

Comprehens i on#2

1 . Due to the high temperature developed as a result

of collision & fusion causes the core of fusion reactor

to plasma.

2 .

3KT 2

2 = 1 2Kq q

r 3KT =

1 2Kq q

r

T =

9

15 5

1.44 10

4 10 3 8.6 10

T = 1.39 × 109 K 9 91 10 T 2 10 K

3 . nt0 = 8 × 1014 × 9 × 10–1 = 7.2 × 1014 > 5 × 1014

nt0 = 4 × 1023 × 1 × 1011 = 4 × 1034 > 5 × 1014

nt0 = 1 × 1024 × 4 × 1012 = 4 × 1036 > 5 × 1014

Comprehens i on#3

1 . E = 2p

2m.... (i) p =

h... (ii)

By equation (i) and (ii)

E =

2

2

h

2m

2 2

2

h (n )

2m(4a )

[ n

2 = a for stationary wave

on string fixed at both end] E a–2

2.

2 –34 2

2 30 9 2

h (6.6 10 ) 1E

e2m4a 2(1.0 10 ) 4 (6.6 10 )

E = 8 meV

3 . mv =

h v =

h

m

hn

m(2a)v n

Comprehension 4

1 . 2 2 2 2

2

n 2

1 I nh / 2 n hE I

2 2I 2I 8 I

2 . 2

2 1 2

hh E E h 4 1

8 I

2

2

hh 4 1

8 I

3446 2

22 11

3h 3 2 10I 1.87 10 kgm

488 10

3 . Moment of inertia of CO molecule about centre of

mass : 2 1 2

1 2

m mI r where =

m m

46

271 2

1 2

I I 1.87 10r

m m 12 16 5 / 3 10

m m 12 16

101.3 10 m

Paragraph 5

1 . 210 206 484 82 2Po Pb He Q

Total energy released = 2Po Pb HeM M M C

= [(209.982876) – (205.974455 + 4.002603)] ×

932 MeV

= [0.005818] × 932 MeV = 5.422376 MeV

Kinetic energy of particle =

A 4 206Q 5.422376 MeV

A 210

= 5.319

MeV = 5319 KeV

2 . Only in option (C); sum of masses of product is less

then sum of mases of reactant

for reaction 2 4 61 2 3H He Li . As M

Li < M

Deutron +

Malpha

Sub jec t i ve

1 . (i) Let at time t, number of radioactive nuclei are N.

Net rate of formation of nuclei of A

dN

dt=–N

dN

N =dt

0

N t

N 0

dNdt

N

Solving this equation, we get

N = 1

[–(–N

0)e–t]...(i)

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(ii) (i) Substituting = 2N0 and t =t

1/2 =

n(2)

in

equation (i) we get, N = 3

2N

0

(ii) Substituting = 2N0

and t in equation (i), we get

N =

= 2N

0 N = 2N

0

2 . Given work function W = 1.9 eV

Wavelength of incident light, = 400 nm

Energy of incident light, hc

E 3.1eV

(Substituting the values of h, c and )

Therefore, maximum kinetic energy ofphotoelectron K

max = E – W = (3.1 – 1.9) = 1.2 eV

Now the situation is as shown below :

e–

K =1.2eVmax

-particles He in fourthexcited state or

n=5(Z=2)

+

He+

n=5E = –2.1eV5

Energy of electron in 4th excited state of He+ (n=5)

will be 2

5 2

ZE 13.6 eV

n

E5= –(13.6)

2

2

(2)

(5)=2.2 eV

Therefore, energy released during the combination

= 1.2 – (–2.2) = 3.4 eV

Similarly energies in other energy states of He+ will

be E4 = –13.6

2

2

(2)

(4)= –3.4 eV

E3 = –13.6

2

2

(2)

(3)=–6.04 eV

E2 =–13.6

2

2

(2)13.6eV

2

The possible transitions are

E54

= E5 – E

4 = 1.3 eV < 2 eV

E53

= E5 – E

3 = 3.84 eV

E52

= E5 – E

2 = 11.5 eV > 4 eV

E43

= E4 – E

3 = 2.64 eV

E42

= E4 – E

2 = 10.2 eV > 4 eV

Hence, the energy of emitted photons in the rangeof 2eV and 4eV are

3.3 eV during combination and

3.84 eV and 2.64 after combination.

3 . Let ground state energy (in eV) be E1

Then, from the given condition E2n

– E1 = 204 eV

1

12

EE 204eV

4n

1 2

1E '1

4n

= 204 eV...(i)

and E2n

– En = 40.8 eV

1 1

2 2

E E

4n n = 40.8 eV

1 2

3E

4n

= 40.8 eV...(ii)

From equation (i) and (ii), 2

2

11

4n 53

4n

2 2

1 151

4n 4n 2

41

n n=2

From equation number (ii),

1

4E

3 n2(40.8) eV = –

4

3(2)2 (40.8) eV

E1 = –217.6 eV E

1 = – (13.6)Z2

Z2 = 1E

13.6

217.616

13.6

Z=4

Emin

= E2n

– E2n–1

1 1

2 2

E E2

4n (2n 1)

1 1E E

16 9 1

7E

144

7

144

(–217.6) eV

Emin

= 10.58 eV

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4 . Energy of incident photon,

Ei=10.6 eV=10.6× 1.6× 10–19J = 16.96 × 10–19 J

Energy incident per unit area per unit (intensity) = 2J

No. of photons incident on unit area in unit time

19

2

16.96 10

1.18 × 1018

Therefore, number of photons incident per unit timeon given area (1.0 × 10–4 m2)

= (1.18 × 1018) (1.0 × 10–4) = 1.18 × 1014

But only 0.83% of incident photons emitphotoelectrons

No. of photoelectrons emitted per second (n)

14

0.53n (1.18 10 )

100

n = 6.25 × 1011

Kmin

= 0 and Kmax

= Ei – work function

= (10.6 – 5.6) eV = 5.0 eV

Kmax

= 5.0 eV

5.(i) Let at time t=t, number of nuclei of Y and Z are NY

and NZ. Then Rate equations of the population of

X,Y and Z are

XdN

dt

=–

X N

X...(i)

YdN

dt

=

X N

X –

YN

Y...(ii)

ZdN

dt

=

YN

Y...(iii)

(ii) Given NY(t) = Y X

0 X t t

X Y

Ne e

For NY to be maximum

YdN (t)0

dt

i.e. X N

X =

Y N

Y... (iv) (from equation (ii))

X Y X0 Xt t t

X 0 YX Y

N(N e ) [e e ]

Y

X

tX Y

tY

e1

e

X Y

X ( ) t

Y

e

(X –

Y)t n (e) = n

X

Y

X

X Y Y

1t n

Substituting the values of X and

Y, we have

1 0.1t n 15 n(3)

(0.1 1 / 30) 1 / 30

t=16.48 s.

(iii) The population of X at this moment,

NX = N

0e–Xt = (1020)e–(0.1) (16.48) N

X = 1.92 × 1019

NY =

X X

Y

N

[From equation (iv)]

= (1.92 × 1019) (0.1)

(1 / 30) = 5.76 × 1019

NZ = N

0– N

X – N

Y = 1020 – 1.92 × 1019 – 5.76 ×

1019

NZ = 2.32 × 1019

6.(i) Given mass of – particle, m=4.002 amu and massof daughter nucleus M=223.610 amu,

de–Broglie wavelength of –particle,

=5.76 × 10–15 m

So, momentum of –particle would be

34

15

h 6.63 10p

5.76 10

kg–m/s

p= 1.151 × 10–19 kg–m/s

From law of conservation of linear momentum, thisshould also be equal to the linear momentum thedaughter nucleus (in opposite direction).

Let K1 and K

2 be the kinetic energies of –particle

and daughter nucleus. Then total kinetic energy in

the final state is: K = K1 + K

2 =

2 2p p

2m 2M

2p 1 1

2 m M

=

2p M m

2 Mm

1 amu = 1.67 × 10–27 kg

Substituting the values, we get

19 2(1.151 10 ) M mK

2 M m

2 27

27 27

p (4.002 223.610)(1.67 10 )

2 (4.002 1.67 10 )(223.61 1.67 10 )

1212

13

10K 10 J

1.6 10

MeV = 6.25 MeV

K=6.25 MeV

(ii) Mass defect, m=6.25

931.470amu = 0.0067 amu

Therefore, mass of parent nucleus

= mass of –particle + mass of daughter

nucleus+mass defect (m)

= (4.002 + 223.610 + 0.0067)amu = 227.62 amu

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7 . The reactor produces 1000 MW power or 109 W

power of 109 J/s of power. The reactor is to function

for 10yr. Therefore, total energy which the reactor

will supply in 10yr is E = (power) (time)

= (109 J/s) (10 × 365 × 24 × 3600 s) = 3.1536 × 1017 J

But since the efficiency of the reactor is only 10%,

therefore actual energy needed is 10 times of it

or 3.1536 × 1018 J. One uranium atom

liberates 200MeV of energy or 200 × 1.6 × 10–13 or

3.2 × 10–11 J of energy. So, number of uranium

atoms needed are

n =

18

11

3.1536 10

3.2 10

= 0.9855 × 1029

or number of kg–moles of uranium needed are

29

26

0.9855 10n 163.7

6.02 10

Hence, total mass of uranium required is

m = (n)M = (163.7) (235) kg

m 38470 kg m = 3.847 × 104 kg

8.(i) Total 6 lines are emit ted.

Therefore n(n 1)

62

n=4

So, transition is taking place between mth energy

state and (m + 3)th energy state. Em =–0.85eV

–13.62

2

Z

m

=–0.85

Z

m= 0.25...(i)

Similarly Em+3

= –0.544 eV

2

2

z13.6 0.544

(m 3)

z0.2

(m 3)

..(ii)

Solving equation (i) and (ii) for z and m.

We get m = 12 and z=3

(ii) Smallest wavelength corresponds to maximum

difference of energies which is obviously Em+3

– Em.

Emax

= – 0.544 – (–0.85) = 0.306 eV

min

= max

hc 1240

E 0.306

= 4052.3 nm.

9 . Area of plates A = 5 × 10–4 m2

distance between the plates d = 1cm = 10–2m

(i) Number of photoelectrons emitted upto t=10 s are

6

(No. of photons falling

in unit area in unit time)n (area time)

10

= 6

1

10[(10)16 × (5 × 10–4) × (10)] = 5.0 × 107

(ii) At time t=10s charge on plate A,

qA=+ne=(5.0 × 107) (1.6 × 10–19) = 8.0 × 10–12 C

and charge on plate B,

qB = (33.7 × 10–12 – 8.0 × 10–12) = 25.7 × 10–12 C

Electric field between the plates E = B A

0

(q q )

2A

E =

123

4 12

(25.7 8.0) 10 N2 10

C2 (5 10 )(8.85 10 )

(iii) Energy of photoelectrons at plate A

= E – W = (5–2) eV = 3eV

Increase in energy of photoelectrons

= (eEd) joule = (Ed) eV

= (2 × 103) (10–2) eV = 20eV

Energy of photoelectrons at plate B

= (20 + 3)eV = 23 eV

1 0 . E = h = Rhc (Z–b)2 2 21 2

1 1

n n

For K– series, b=1 = Rc(Z–1)2 2 21 2

1 1

n n

Substituting the values,

4.2 × 1018 = (1.1 × 107) (3 × 108)(Z–1)21 1

1 4

(Z–1)2 = 1697 Z –1 41 Z = 42

1 1 . Maximum kinetic energy of the photoelectrons

would be Kmax

= E – W = (5 – 3) eV = 2eV

Therefore, the stopping potential is 2V. Saturation

current depends on the intensity of light incident.

When the intensity is doubled the saturation current

will also become two fold. The corresponding graphs

are shown in figure.

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1 2 . Let n0 be the number of radioactive nuclei at time

t=0. Number of nuclei decayed in time t are given

by n0(1–e–t), which is also equal to the number of

beta particles emitted during the same interval of

time. For the given condition,

n = n0(1–e–2) ...(i) (n + 0.75n) = n

0(1–e–4) ...(ii)

Dividing equation (ii) by (i), we get

1.75=

4

2

1 e

1 e

1.75 – 1.75 e–2 = 1 – e–4

1.75 e–2 – e–4 = 3

4...(iii)

Let us take e–2 = x

Then the above equation is x2 – 1.75 x + 0.75 = 0

21.75 (1.75) (4)(0.75)

x2

x=1 and

3

4

From equation (iii) either e–2 = 1, e–2 = 3

4

but e–2 = 1 is not accepted because which means

=0. Hence, e–2 = 3

4

– 2 n(e) = n (3) – n(4) = n(3) – 2 n(2)

= n (2) –1

2n(3)

Substituting the given values,

= 0.6931 – 1

2 × (1.0986) = 0.14395 s–1

Mean–life tmean

= 1

= 6.947 s

1 3 . Wavelengths corresponding to minimum wavelength

(min

) or maximum energy will emit photoelectrons

having maximum kinetic energy.

(min

) belonging to Balmer series and lying in the

given range (450 nm to 750 nm) corresponds to

transition from (n=4 to n=2). Here,

E4 = – 2

13.60.85eV

(4) E

2 = 2

13.63.4eV

(2)

E = E4 – E

2 = 2.55 eV

Kmax

= Energy of photon – work function

= 2.55 – 2.0 = 0.55 eV

1 4 . Let N0 be the initial number of nuclei of 238U.

After time t, NU = N

0

n1

2

Here n = number of half–lives

=

9

91 / 2

t 1.5 10 1

t 34.5 10

N

U = N

0

1

31

2

and NPb

= N0 – N

U = N

0

1 / 31

12

1 / 3

U

3Pb

1N 2 3.861N 1

12

1 5 . (i) From the relation r A1/3

We have

1 / 31 1

2 2

r A

r A

1 / 32 1 / 3

A(14)

4

A2 = 56

(ii) Z2 = A

2 – no. of neutrons = 56 – 30 = 26

fka = Rc (Z–1)2 2 2

1 1 3Rc

41 2

(Z–1)2

Substituting the given values of R, c and Z.

We get fka = 1.55 × 1018 Hz

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1 6 . For 0 < x < 1, PE = E0

Kinetic energy K1 = Total energy – PE

= 2E0 – E

0 = E

0 1

0

h

2m(2E ) ...(i)

For x > 1, PE = 0

Kinetic energy K2 = Total energy =2E

0

2

0

h

2m(2E ) ...(ii)

From equation (i) and (ii), we have1

2

2

1 7 .

P

P P

hp 2mqV

m q4 2 8 3

m q

1 8 .t

0

dNN e

dt 0

dNn n N t

dt

1

0

dN 1n n N t slope year

dt 2

11

yr2

1 / 2

0.693t 1.386

given time is 3 times of t

1/2

value of p is 8.

1 9 . eVs = h – S

hV

e e

vs

slope =h

e= constant ratio = 1

2 0 . Given T½ = 1386 sec.

3dN

10dt

Fraction

=

tt0 0 0

0 0

N N N N e1 e

N N

= ln 2

8013861 e

=

4

1004

1 e 1 1100

=

44%

100