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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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