Phy 1 (16)

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w

g

g

w

g

w

n

n

n/C

n/C

v

v

Since frequency f of light does not change, vw = wf and vg = gf we obtain,

w

g

g

w

g

w

n

n

v

v

8

9

3/4

2/3

g

w

8

1

8

89

g

gw

8

1

g

.

Q.5. Momentum of a photon = h/

momentum of electron = mvso = h / mv

=531

34

102101.9

1063.6

= 3.6 nm

Q.6. 2/12/1 T/t)T(

0 2

1

e

1

N

N

also4

1

N

N

0

1 or t1 = 2T1/2

&81

NN

0

2 or t2 = 3T1/2

t = 10 = t2 – t1 = T1/2 T1/2 = 10 sec.

Tmean =693.0

T 2/1 = 14.43 sec.

Q.7. = a(z - b)

)bz(ac

11

. . . . (i)

& )bz(ac 22

. . . (ii)

From (i) - (ii),

21

11c = a(z1 - z2 )

a = 5 107 (Hz)1/2

From (i) /(ii),bz

bz

2

1

1

2

b = 1.37

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Q.8. Removal of an electron results into He+, and energy required to remove an electron from He+ = z2

13.6 eV

= 4 13.6 = 54.4 eVso total energy required= 54.4 + 24.6

= 79 eV.

Q.9. According to Bohr theory

22

21

2

n

1

n

1RZ

1

Thus2Z

1

More the atomic number, smaller is the wavelength obtained in the transition of electron (for

identical transition).

Q.10. )bZ(Rv

or 2)1Z(1

(If b is small)

2

si

u

91

46

u = 0.146 A°

Q.11. E2 – E1 = hc

2 1

1 1

= 1 21 2

hc( )

taking approximation that

if y 2 then

1/ 21 21 2( )

2

E2 – E1 = 2 10-3 eV.

Q.12.

1 2 3

hc hc hc

3 = 1 21 2

.

Q.13. dN

n Ndt

0

N t

N 0

dNdt

n N


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0n N1 lnn N

= t

N = t0

n nN e

Q.14. Current is independent of work function of metal. It depends on photon density onlyhence I1= I2

Q.15. Let at time ‘t’ number of radioactive nuclear be N.

Net rate of formation of nuclear of A isdN

dt = – N

Or,dN

N = dt

orN

N0

dN

N =t

0dt this gives N = t0

1N e

Q.16. 1

2

/hc

/hc

2

1

= 1.05 eV

Q.17. )bZ(Rv

or 2)1Z(1

(If b is small)

2

si

u

91

46

u = 0.146 A°

Q.18. 1

2

/hc

/hc

2

1

= 1.05 eV

Q.19. Q = total energy released

=2

1mv

2 +2

1 myvy

2 . . . . . (i)

by conservation of momentum

mv + myvy = 0 . . . . . (ii)

from equation (i) and (ii)

Q =2

1 mv

2

ym

m1

Q = 5.3410 MeValso Q = (Mx – my m) 931.5 Mx = 239.048 amu.

Q.20. Velocity of neutrons =27

19

n 1067.1

0327.0106.12

m

eV2


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= s/m1029.6 6 = 2.5 103 m/s

Time taken by neutron to cover the distance 10 m

t = 33

104105.2

10

seconds

Now t

0

eN

N

where N is number of neutron after t seconds i.e. fraction of neutron decayed after t seconds will be

1-4109.93104

0

eN

N

sec/109.9700

693.0 4 = 4 10-6

Q.21. According to Moseley’s equation for k radiation ;

222

21

11)1z(R1

5405.1

7092.0

)1z(

)1z(2

1

21

z = 29 for cu, hence z1 – 1 = 286578.1

5405.1 = 27

or z1 = 4z

Impurity is molybdenum

similarly ;5405.1

65768.1

)1z(

)1z(2

2

2

1

2

or z2 – 1 = 28 276578.1

5405.1

z2 = 28

It is atomic number of Nickel.

Hence the other impurity is Nickel.

Q.22. Energy of a photon = E =6000

12400hc

= 206 eV

= 2.06 1.6 10-19 J= 3.3 10-19 J

Photon flux =19103.311400

EIA

= 4.22 1021 photon/sec.

Q.23. Momentum of the photon , p= E

c

133 10 1 6 10

3 10

6 19

8

. .

= 7.0910-22 kgms-1 Using momentum conservation recoil energy of the


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60 Ni is E =

p

m

2 22

2

272

7 09 10

2 60 167 10

.

.= 2.5110-18 J = 1.57 10-5 MeV

Recoil speed of the nucleus, v = pm

m s

7 09 10

60 1 67 107 07 10

22

27

3.

.. /

Q.24. (i) Energy of each photon = E =

hc = 3.975 10-19 J

No. of photons falling on surface per second & being absorbed,

n=eV48.2

J10 = 2.52 1019 eV

(ii) The linear momentum of each photon = p =c

hh

Total momentum of all photons (falling in one sec.)

=8

nh 10J

c 3 10

= 3.33 10-8 N-s

Rate of change of momentum = Force =dt

dp = 3.33 10-8 N.

Q.25. By relation,

eV =

hc, where is work function

Differentiating w.r.t. , we get

2

hc

d

dVe

or dV = -

de

hc2

dv = -)106.1()104000(

10210310623.619210

10834

= - 1.56 mV.

Q.26. Moseley’s law

22

n

111zR

1 for K- lines where n =2,3,4,.....................

For K-absorption edge

R

11z

or z =

74110097.110171.0

1710

The element is Tungsten.

Q.27. Current is independent of work function of metal. It depends on photon density only hence


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2

1

I

I = 1

Q.28. Velocity of neutrons = 27

19

n 1067.1

0327.0106.12

m

eV2

= s/m1029.6 6 = 2.5 103 m/s

Time taken by neutron to cover the distance 10 m

t = 33

104105.2

10

seconds

Now t

0

eN

N

where N is number of neutron after t seconds i.e. fraction of neutron decayed after t seconds will be

1-4

109.93

104

0e

NN

sec/109.9700

693.0 4 = 4 10-6

Q.29. At time t lets say there are N atoms. In time dt, dN1 decays and dN2 produces

dN2 =1000106.1

dt1019

4

dN1 = N dt.

Total production in time dt dN =

N

106.1

112

dt

3600

0

N

011

dtN1025.6

dN0

- 36001025.6

N1025.6log

111

011

N0 = 1.8 108 . (given) = 8 10-8

t1/2 = 8108

6931.0

= 8.66 106 sec.

= 100.26 days.

Q.30.

2221

2

n

1

n

1

RZ

1

2227 n

1

2

14109678

10121

1

n2 = 4

no. of revolution per second =ncecircumfere

orbitthatinvelocity

f = 02

0

v n / z

n2 r

z

= 4.09 1014 Hz.


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Q.31. En – E1 =

hc

after putting the values we getn = 4

possible lines in the resulting emission spectrum = 6.

Q.32.

t

12

2t

12

1

0

3 12 ee1N

N 1 =

30

6931.0 2 =

45

6931.0, t = 60 min.

= [1 + (-3) (2)-4/3 - (-2) (2)-2 ]

= [1 - 3 (2)-4/3 + 2(2)-2 ]=0.71

Q.33. E =

hC = + (KE)max

3.1 = 2.5 + (KE)max

(KE)max = 0.6 ev = 0.6 1.6 10 –19 J = 9.6 10 –20 Jp = mK2 = 2031 106.9101.92 = 4.2 10 –25 kg – m/sec

Q.34. (a)2

2613

n

Z.En

Excitation energy = E = E3 - E1 = -13.6 (3)2

22 1

1

3

1

= +13.6 (9) [1 - 1/9] = 13.6 (9) (8/9) = 108.8 eV.

19834

106.18.108

1031063.6

E Δ

hc

Wavelength

= 114.3 A

(b) From the excited state (E3), coming back to ground state, there can be3C2 = 3 possible

radiations.

Q.35. Mass defect

m = (2 2.0141 – 4.0026) amuor m = (2 2.0141 – 4.0026) 931 MeVEnergy used in reactor per reaction

=100

25 (2 2.0141 – 4.0026) 931 = 5.9584 MeV

= 9.5334 10-13 Joule.

Total energy obtained per day

= (200) MW 24 60 60 sec.

Mass of deuterium required

=13

621

105334.9

)60602410200)(106691.0(

= 120 g.


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Q.36. dt

dNq - N

q = rate of production of nucleiN = number of nuclei in radionuclide at any instant

= decay constant = T2ln

T = half life

dt

dNq - N = q -

T

2lnN

T

2lnNqT

dt

dN

T

dt

2lnNqT

dN

. . . (1)

integrating equation (1)

t

0

N

0

dt

T

1

2lnNqT

dN

N =

T

2lnt

e12ln

qT =

1620

2ln3240

e12ln

16201000 = 1.753 106

Hence, rate of decay A = N = q

T

2lnt

e1

Hence rate of release of energy at this time,

= AE0 = qE0 (1 - etln2/T) = 1000 200 (1 – e(3240 ln2/1620))

= 150 103 MeV/sec.Total number of nuclei decayed upto this time = q t – Nhence total energy released upto this time

= (qt – N) E0 = 297.43 106 MeV.

Q.37. Let at time ‘t’ number of radioactive nuclear be N.

Net rate of formation of nuclear of A isdt

dN = – N

Or,N

dN

= dt

or N

0N N

dN =

t

0dt this gives N = t0 eN

1

Q.38. (a) E =

hC

= + (KE)max

400

1242 = + (KE)max

3.1 = 2.5 + (KE)max (KE)max = 0.6 ev = 0.6 1.6 10

–19 J = 9.6 –20 J

p = mK2 = 2031 106.9101.92 = 4.2 10 –25 kg – m/sec

(b) Stopping potential

e

KE max = 0.6 volt


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Q.39. Let be decay constant

Ndt

dN = Rate of formation of nuclei

dt

N

dN

Integrating both side we get

ln ( - n

0)N = - t

i.e. ln tN

(at t = 0, N = 0)

teN

i.e. N =

)e1( t

so as t , e-t

0so after very long time

N = ttancons

or N =693.0

T 2/1

Q.40. The maximum energy of photon during de-excitation will be if transition takes place between (2n)states to ground state.

hence, 204 = 13.6

2n4

1

1

1z2 … (i)

also 40.8 = 13.6 222

zn41

n1

… (ii)

dividing (i) by (ii) we get

5 =3

1n4 2 n = 2

substituting n = 2 in equation (i)

15 =16

15z2

z2 = 16 or z = 4The minimum energy during de-excitation will be if transition takes place between two outer mostadjacent orbit. i.e. n = 4 to 3.

Emin = 13.6 222 44

1

3

1

= 10.57 eV

Q.41. Mass defect m = 2mD – mT – mP = 2 2.01458 – 3.01605 – 1.00728= 0.00583 amu.

hence E = 0.00583 930 1.6 10-19 106 = 8.675 10-13 J

The efficiency of the retardation is 60 %


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Eavailable = 0.6 8.675 10-13 J

= 5.205 10-13 JTotal energy required in one hour

E = 108 3600 = 36 1010 JHence number of deuterium nuclides required

= EE2 1.38 1024

Q.42. (i) A = 228 + 4 = 232

& 92 = z + 2 z = 90

(ii) Bqvr

vm 2

v = 1.59 10

7 m/s

From COM, m v = my vy Thus, energy released or the sum of kinetic energies of the products

=

2

1

y

222

m

vmvm

= 5.342 MeV= 0.0057 amu.

Applying COE,

Mass of 23292 X = my + m + 0.0057 amu.

= 232.0387 amu

Mass defect = 92 (1.008) + 140 (1.009) - 232.0387= 1.9573 amu = 1823 MeV

Q.43. Reactant Products

1023 Ne 22.9945-10me 1123 Na 22.9898-11me

10 me 0 .

Total 22.9945-10me 22.9898-10me

mass defect = 22.9945-22.9898=0.0047u

Since, 1u = 931.4 MeV

Energy release = (0.0047)(931.4)= 4.4MeV

The major portions of this energy is shared by particle and anti-neutrino. Hence the energy rangeof

- particle varies from 0 to 4.4 Mev.

Q.44. h1 = 13.6 (2)2 1.6 10-19

22 )1n(

1

n

1

hf = 13.6 (2)2 1.6 10-19

1

n

12

h(f - 1) = 2

192

)1n(

106.1)2(6.13

f - 1 = 3.3 10

15 (given)

(n + 1)2 = 4 n = 1


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1

hc 13.6 (2)2 1.6 10-19

4

1

1

1 .

1 = 303 A0.

Q.45. At time t lets say there are N atoms. In time dt, dN1 decays and dN2 produces

dN2 =1000106.1

dt1019

4

dN1 = N dt.

Total production in time dt dN =

N

106.1

112

dt

0N 3600

11

0 0

dNdt

6.25 10 N

-11

0

11

6.25 10 N1ln 3600

6.25 10

N0 = 1.8 108 . (given)

= 8 10-8

t1/2 = 8108

6931.0

= 8.66 106 sec.

= 100.25 days.

Q.46. Mass diff. = 2 2.015 – (3.017 + 1.009) = 0.004 amu. energy released = 0.004 331 MeV = 3.724 MeV

energy released per deuteron = 724.32

1 MeV

No. of deuteron in 1 kg =

2

1002.6 26

energy released / kg = 1.862 0.301 1026 MeV 9 1013 J.

Q.47. (i). Energy of each photon = E =

hc = 3.975 10-19 J

No. of photons falling on surface per second & being absorbed,

n=eV48.2

J10 = 2.52 1019 eV

(ii). A = 228 + 4 = 232

& 92 = z + 2 z = 90

(1.v) A radioactive nuclide with half life period T is produced at the constant rate of n per second.

The number of radioactive nuclide at t = 0 is N0 , find

(i) the number of radioactive present at time t

(ii) the maximum number of these radioactive nuclei

(1.v) Nndt

dN

t

0

Nt

0N

dtNn

dN

t0

t

eNnnN

Where N is maximum, 0dt

dN N =

n


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Q.48. (a)2

0

22

r 4

Ze

r

mv

. . . (i)

& mvr =

2

nh . . . (ii)

From (i) & (ii), r =2

022

meZ

hn

Here z = 3 & m = 208 me, r = 2

e

022

em624

hn

,

(b)2

e

02

2e

022

em

h

em624

hn

n 25

(c) En = Total Energy = - 2220

42

0

2

hn8

mez

r 8

ze

=

220

4e

2 h8

em

n

1872

= - 6.13n

18722

eV

E1 = -25.4 keV ; E3 = -2.8 keV & E3 - E1 = 22.6 keV .

Required wavelength = =E

hc

= 55 pm.

Q.49. Mass defect

m = (2 2.0141 – 4.0026) amuor m = (2 2.0141 – 4.0026) 931 MeV

Energy used in reactor per reaction

=100

25 (2 2.0141 – 4.0026) 931 = 5.9584 MeV

= 9.5334 10-13 Joule.

Total energy obtained per day

= (200) MW 24 60 60 sec.

Mass of deuterium required

=13

621

105334.9

)60602410200)(106691.0(

= 121 g.

Q.50. For photons of 1 = 4000A, energy E1 =4000

12375eV = 3.094 eV

and for 2= 6000A, energy E2 =6000

12375eV = 2.061 eV

Thus photo-electronic emission is possible with 1 only, which experience Lorenz force and movealong circular path. The ammeter will indicate zero deflection if the photoelectrons just completesemi-circular path before reaching the plate P. Thus separation d = 2r = 10 cm

r = 5m


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but, r =qB

mv Bmin =

qr

mv

Now, eV39.2094.3Whc

mv2

1

1

2

Substituting the value for v, m, q and r,Bmin = 5.6610

-5T

Q.51. de-Broglie wavelength 1 =p

h =

mve2

h

Shortest X-ray wavelength 2 =ve

hc

m2

ve

c

1

mve2c

ve

2

1

Substituting the values, 113

8

2

1 108.12

1010

103

1

= 0.1

Q.52. (a) Using Einstein’s relationEmax = hf - Wo here Emax = hf - Wo = 13.6 eV

and Emax =

19883

o106.11015.12

1031063.6hcWf

6

5h

= 10.2 eV

From the above two equations,

6

hf 3.4 eV

or f =

34

19

1063.6

106.14.36

= 4.92 1015 Hz

(b) Wo = hf - 13.6 = 6(3.4) - 13.6 = 6.8 eV

Q.53. (i) E = +2

1m 2maxv

= 1.82 + 0.73 = 2.55 eV

(ii) En = -2n

6.13eV (for hydrogen atom)

E1 = -13.6 eV, E2 = - 3.4 eV, E3 = -1.51 eV, E4 = -0.85 eV

clearly E4 – E2 = -0.85 – (-3.4) = 2.55 eV.Hence quantum levels involved are 4 and 2.

(iii) = n

2

h

change in angular momentum = 4 - 2

= (4 – 2)

h

2

h


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(iv) If P = linear momentum, then

P =c

E = 1.36 10-27 kg ms-1

If v = recoil speed of hydrogen atom of mass M, then from COM,Mv = P

or v =M

P = 0.85 ms-1.

Q.54. (a) Number of photoelectrons emitted from plate A upto

t =10 s 76

164

e 1051010

10)105(n

(b) Charge on plate B at t = 10 sec

Qb = 33.7 10-12 – 5 107 1.6 10-19

= 25.7 10-12 Calso Qa = 8 10

-12 C

)QQ( A2

1

22E AB

00

A

0

B

=124

12

1085.8105

107.17

= 2000 N/C

A B

d =1 cm

(c) K.E. of most energetic particles

= (h -) + e(Ed) = 23 Ev

Q.55. 1102

m10p

11

1500

11

…. (i)

or E = J10

P

11

1500

hchc 102

=

219

10

p

11

106.0(1500

10hceV

= 8.28

2p

11 eV

(a) wavelength of the most energetic photon is corresponding to p = , = 1500 10-10 m = 1500 A0 wavelength of least energetic photon. corresponding to

p = 2, hence again

= 1500 10-10 14

4

= 2000 A0

(b) According to equation

E = 8.28

)P1(

P28.8

P1P

2

22

2

= eV

n28.8 2

or En = - 2n

28.8eV

for n = 1, E1 = - 8.28 eV n = 3_______ - 0.92 eV

for n = 2, E2 = - 2.07 eV n = 2_______ -2.07 eV

for n = 3, E3 = -0.92 eV n = 1_______ -8.28 eV

(c) Ionization energy = - E1 = 8.28 eV


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Hence ; ionization potential = 0.28 volt.

Q.56. (a)2

0

22

r 4

Ze

r

mv

. . . (i)

& mvr =

2

nh . . . (ii)

From (i) & (ii), r =20

22

meZ

hn

Here z = 3 & m = 208 me,

r = 2e

022

em624

hn

,

(b)2

e

02

2e

022

em

h

em624

hn

n 25

(c) En = Total Energy = - 2220

42

0

2

hn8

mez

r 8

ze

=

220

4e

2 h8

em

n

1872

= - 6.13n

18722

eV

E1 = -25.4 keV ; E3 = -2.8 keV& E3 - E1 = 22.6 keV .

Required wavelength = =E

hc

= 55 pm.

Q.57. = a(z - b)

)bz(ac

11

. . . . (i)

& )bz(ac

22

. . . . (ii)

From (i) - (ii),

21

11c = a(z1 - z2 )

a = 5 107 (Hz)1/2

From (i) /(ii),

bz

bz

2

1

1

2

b = 1.37

Q.58. Q value of the reaction, Q = (2 4 7.06 – 7 5.6) MeVQ = 17.28 Mev

Applying C.O.E for collision

kp + Q = 2 k . . . (i)

ppkm2 = 2 km2 cos

kp = 16 k cos2

kp = k ( cos = ¼)

Li p


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Putting in (i) we getkp = Q = 17.28 MeV

Q.59. The quantum number of the initial state is given by n, where2

)1n(n = 15, i.e. n = 6

The work function of the photo cathode (B) is given by

E=199

834

106.110830

1031063.6

eV

= 1.5 eV.

The maximum velocity of the emitted photoelectrons is given by

v =3

2

0

0

10

107.3

B

E = 3.7 105 m/s

the kinetic energy of the fastest photoelectrons is

Emax = 19

25312

106.1

)107.3(101.9

2

1mv

2

1

eV

= 0.39 eVThe energy of the emitted photons is

1.5 eV + 0.39 eV = 1.89 eVIf the atomic number of the atoms is Z, and the quantum number of the final state is m, then

13.6 Z2 89.16

1

m

122

Rewriting this equation in the form:

z =22

22

22

m6

m6

7

1

6

1

m

1

6.13/89.1

or, z = )m36(7

m6

2

We get by substituting possible values of m : 5, 4, 3 …… etc ; for the only possible integral value ofz.The correct values are: n = 6, m = 4 and z = 2.

Q.60. (a) Radius of circular path r =qB

mk2

qB

mv

k =m2

)rqB( 2 = 0.86 eV

(b) E3 – E2 = 13.6 (1/22 – 1/32) = 1.9 eV

= h - (KE)max = 1.04 eV

(c) E = hc/

=19

348

106.19.1

106.6103

= 6513 A0.


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Q.61. (a) The wavelength of emitted radiation is given as

1

= z2R

22

21 n

1

n

1

for the Lithium atom z = 3

1 = 32R

22 4

1

3

1 = 9R

144

7

= 2.08 10-7 m(b) Relation orbit of the electron

r =qB

mv

v =m

qBr = 1.18 106 m/s

KE of the electron =2

1 mv2 = 3.96 eV

(c) KEmax =hc -

=

hc - KEmax = 2eV

Q.62. a) n=4 --------------------- E4 = 1.125 eVn=3 --------------------- E3 = 2.0 eV

n=2 --------------------- E2 = 4.5 eV

n=1 --------------------- E1 = 18eV

b) Excitation potential for state n = 2 is 18-4.5 = 13.5 V

c) Energy of the electron accelerated through a potential difference of 16.2 V is 16.2eV At the most, it can excite electron fron n=1 to n=3The number of possible wavelength are 3

22

21

19

n

1

n

1

hc

106.1181x

For transiton 3-2, n1 =2, n2 = 3.

32 = 4970 A

For 3-1; n1 = 1;n2 =3

31 = 777 A

For 2-1;n1 = 1;n2 = 2

21 = 920 A

d) No

The energy corresponding to = 2000 A

is


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E =

1478

106.1102

1031063.6hc34

= 6.21 eV

The minimum excitation energy is 13.5 eV.

e) Minimum photoelectric wavelength is

min = 690106.118

hc19

A

Q.63. The nucleus reaction can be written as

n1 + C12 Be9 + He4 Q value of the reaction

= - Tth

projectiletheof massetargttheof massetargttheof mass

= - 6.17

12112

MeV

n0Be

He4

[k1, k3, k4 [mn, mBe mHe is the kinetic energy and masses of n1, Be9, He4 then corresponding

momentum is 1nkm2 , 3Bekm2 , 4Hekm2

Conservation of momentum

1nkm2 = 3Bekm2 cos . . . (i)

4Hekm2 = 3Bekm2 sin . . . (ii)

From (i) and (ii) we getmn k1 + mHe k4 = mBe k3 . . . . (iii)nn = 1 mHe = 4 mBe = 9k1 + 4k4 = 9k3 . . . . . (iv)conservation of energyk1 + Q = k3 + k4 . . . . (v)From (iv) and (v) we get,

9

k4k 41 = 41 kQk

k4 =13

Q9k8 1

= 2.2 MeV.

Q.64. (i) E = +2

1 m 2maxv

= 1.82 + 0.73 = 2.55 eV

(ii) En = - 2n

6.13eV (for hydrogen atom)

E1 = -13.6 eV, E2 = - 3.4 eV, E3 = -1.51 eV, E4 = -0.85 eV

clearly E4 – E2 = -0.85 – (-3.4) = 2.55 eV.

Hence quantum levels involved are 4 and 2.


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(iii) = n

2

h

change in angular momentum = 4 - 2

= (4 – 2)

h

2

h

(iv) If P = linear momentum, then

P =c

E = 1.36 10-27 kg ms-1

If v = recoil speed of hydrogen atom of mass M, then from COM,Mv = P

or v =M

P = 0.85 ms-1.

Q.65. (a) For 1 = 5000 A0, E1 =

4500

12400hc

1

eV = 2.75 eV

2 = 6000 A0 E2 =

2

hc

= 2.06 eV

and for 3 = 12000 A0 E3 = 1.03 eV

The energy of excitation of H atom (n = 2 to n = 4)

= 13.6

22 4

1

2

1eV 2.55 eV

The work function of the material = (2.75 – 2.55) eV = 0.2 eVThe maximum energy P.E.'s have energies of 2.55 eV and 2.06 – 0.2 = 1.86 eV and 1.03 – 0.2 =0.83 eV

The de-Broglie wavelengths are9.110

124003 A0 ,

86.110400.123 A

0 ,83.01000

12400

or 8.99A0, 9.09 A0, 13.6 A0

(b) All three wavelengths will cause photo emission.Number of photoelectrons /sec

NPE = 1.44 102

3

1 0.2 1 10-4

03.1106.1

1

06.2106.1

1

75.2106.1

1191919

The photocurrent = 1.74 mA

(c) If the work function was 40 % lower, the third wavelength would also cause photoemission.The stopping potential = (2.75 - 0.2)V = 2.55 V.

Q.66. (a) According to the problem, – decay is given by,92X

A zY

228 +

i.e. 92X A zY

228 + 2He4

A = 228 + 4 = 232Z = 92 – 2 = 90 [1]

(b) Since – particle moves in a circular orbit in the magnetic field


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Bqvr

vm2

v = m

rqB 1.59 107 m/s [1]

From law of conservation of momentummv = myvy

Ey =y

22

m2

vm [1]

So the sought energy E = E + Ey

E =

y

2

y

22

m

m1

2

vm

m

vmvm

2

1 [1]

= 5.342 MeV = 0057.05.931

342.5 amu

Applying the principle of conservation of energy, mass of 92X

232

= my + m + 0.0057 amu= 228.03 + 4.003 + 0.0057 = 232.0387 amuSince 92X

232 contains 92 protons and 140 neutrons , binding energy = mass defect= 92(1.008) + 140(1.009) – 232.0387

= 1.9573 amu = 1.9573 931.5 = 1823 MeV [1]

Q.67. (a)2

2613

n

Z.En

Excitation energy = E = E3 - E1 = -13.6 (3)2

22 1

1

3

1

= +13.6 (9) [1 - 1/9] = 13.6 (9) (8/9) = 108.8 eV.

19

834

106.18.108

1031063.6

E Δ

hcWavelength

= 114.3 A

(b) From the excited state (E3), coming back to ground state, there can be3C2 = 3 possible

radiations.

Q.68. (a) If x is the difference in quantum number of the twostates

x 12then C 6

x = 32

2

z (13.6 eV)Now, we have 0.85eV ...(i)

n

2

2

z (13.6eV)and 0.544eV .. .(ii)

(n 3)

solving (i) and (ii) we getn = 12 and z = 3

n+3

n

Smallest

(b) Smallest wavelength is given byhc

(0.85 0.544)eV

Solving, we get 4052 nm.


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