HC Verma Solutions: Chapter 46 - The Nucleus | Physics Class 11 - NEET PDF Download

 Page 1


46.1
THE NUCLEUS
CHAPTER - 46
1. M = Am
p
, f = M/V, m
p
= 1.007276 u
R = R
0
A
1/3
= 1.1 ? 10
–15
A
1/3
, u = 1.6605402 ? 10
–27
kg
= 
27
3
A 1.007276 1.6605402 10
4 / 3 3.14 R
?
? ? ?
? ?
= 0.300159 ? 10
18
= 3 ? 10
17
kg/m
3
.
‘f’ in CGS = Specific gravity = 3 ? 10
14
.
2. f = 
30
13 14
17
M M 4 10 1 1
V 10 10
v f 0.6 6 2.4 10
?
? ? ? ? ? ? ?
?
V = 4/3 ?R
3
.
?
14
1
10
6
? = 4/3 ? ? R
3
? R
3
= 
14
1 3 1
10
6 4
? ? ?
?
? R
3
= 
12
1 100
10
8
? ?
?
? R = ½ ? 10
4
? 3.17 = 1.585 ? 10
4
m = 15 km.
3. Let the mass of ‘ ?’ particle be xu.
‘ ?’ particle contains 2 protons and 2 neutrons.
? Binding energy = (2 ? 1.007825 u ? 1 ? 1.00866 u – xu)C
2
= 28.2 MeV (given).
? x = 4.0016 u. ?
4. Li
7
+ p ? l + ? + E ; Li
7
= 7.016u
? = 
4
He = 4.0026u ; p = 1.007276 u
E = Li
7
+ P – 2 ? = (7.016 + 1.007276)u – (2 ? 4.0026)u = 0.018076 u.
? 0.018076 ? 931 = 16.828 = 16.83 MeV. ?
5. B = (Zm
p
+ Nm
n
– M)C
2
Z = 79 ; N = 118 ; m
p
= 1.007276u ; M = 196.96 u ; m
n
= 1.008665u
B = [(79 ? 1.007276 + 118 ? 1.008665)u – Mu]c
2
= 198.597274 ? 931 – 196.96 ? 931 = 1524.302094
so, Binding Energy per nucleon = 1524.3 / 197 = 7.737.
6. a) U
238
2
He
4
+ Th
234
E = [M
u
– (N
HC
+ M
Th
)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev.
b) E = U
238
– [Th
234
+ 2n ?
0
+ 2p ?
1
]
= {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u
= 0.024712u = 23.0068 = 23.007 MeV.
7.
223
R
a
= 223.018 u ; 
209
Pb = 208.981 u ; 
14
C = 14.003 u.
223
R
a
?
209
Pb + 
14
C
?m = mass 
223
R
a 
– mass (
209
Pb + 
14
C)
? = 223.018 – (208.981 + 14.003) = 0.034.
Energy = ?M ? u = 0.034 ? 931 = 31.65 Me. ?
8. E
Z.N.
? E
Z–1
, N + P
1
? E
Z.N.
? E
Z–1
, N + 
1
H
1
[As hydrogen has no neutrons but protons only]
?E = (M
Z–1, N
+ N
H
– M
Z,N
)c
2
?
9. E
2
N = E
Z,N–1
+ 
1
0
n .
Energy released = (Initial Mass of nucleus – Final mass of nucleus)c
2
= (M
Z.N–1
+ M
0
– M
ZN
)c
2
.
10. P
32
? S
32
+ 
0 0
0 1
v ? ?
Energy of antineutrino and ?-particle 
= (31.974 – 31.972)u = 0.002 u = 0.002 ? 931 = 1.862 MeV = 1.86.
11. In ? P + e
–
We know : Half life = 0.6931 / ? (Where ? = decay constant).
Or ? = 0.6931 / 14 ?60 = 8.25 ? 10
–4
S [As half life = 14 min = 14 ? 60 sec].
Energy = [M
n
– (M
P
+ M
e
)]u = [(M
nu
– M
pu
) – M
pu
]c
2
= [0.00189u – 511 KeV/c
2
]
= [1293159 ev/c
2
– 511000 ev/c
2
]c
2
= 782159 eV = 782 Kev. ?
Page 2


46.1
THE NUCLEUS
CHAPTER - 46
1. M = Am
p
, f = M/V, m
p
= 1.007276 u
R = R
0
A
1/3
= 1.1 ? 10
–15
A
1/3
, u = 1.6605402 ? 10
–27
kg
= 
27
3
A 1.007276 1.6605402 10
4 / 3 3.14 R
?
? ? ?
? ?
= 0.300159 ? 10
18
= 3 ? 10
17
kg/m
3
.
‘f’ in CGS = Specific gravity = 3 ? 10
14
.
2. f = 
30
13 14
17
M M 4 10 1 1
V 10 10
v f 0.6 6 2.4 10
?
? ? ? ? ? ? ?
?
V = 4/3 ?R
3
.
?
14
1
10
6
? = 4/3 ? ? R
3
? R
3
= 
14
1 3 1
10
6 4
? ? ?
?
? R
3
= 
12
1 100
10
8
? ?
?
? R = ½ ? 10
4
? 3.17 = 1.585 ? 10
4
m = 15 km.
3. Let the mass of ‘ ?’ particle be xu.
‘ ?’ particle contains 2 protons and 2 neutrons.
? Binding energy = (2 ? 1.007825 u ? 1 ? 1.00866 u – xu)C
2
= 28.2 MeV (given).
? x = 4.0016 u. ?
4. Li
7
+ p ? l + ? + E ; Li
7
= 7.016u
? = 
4
He = 4.0026u ; p = 1.007276 u
E = Li
7
+ P – 2 ? = (7.016 + 1.007276)u – (2 ? 4.0026)u = 0.018076 u.
? 0.018076 ? 931 = 16.828 = 16.83 MeV. ?
5. B = (Zm
p
+ Nm
n
– M)C
2
Z = 79 ; N = 118 ; m
p
= 1.007276u ; M = 196.96 u ; m
n
= 1.008665u
B = [(79 ? 1.007276 + 118 ? 1.008665)u – Mu]c
2
= 198.597274 ? 931 – 196.96 ? 931 = 1524.302094
so, Binding Energy per nucleon = 1524.3 / 197 = 7.737.
6. a) U
238
2
He
4
+ Th
234
E = [M
u
– (N
HC
+ M
Th
)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev.
b) E = U
238
– [Th
234
+ 2n ?
0
+ 2p ?
1
]
= {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u
= 0.024712u = 23.0068 = 23.007 MeV.
7.
223
R
a
= 223.018 u ; 
209
Pb = 208.981 u ; 
14
C = 14.003 u.
223
R
a
?
209
Pb + 
14
C
?m = mass 
223
R
a 
– mass (
209
Pb + 
14
C)
? = 223.018 – (208.981 + 14.003) = 0.034.
Energy = ?M ? u = 0.034 ? 931 = 31.65 Me. ?
8. E
Z.N.
? E
Z–1
, N + P
1
? E
Z.N.
? E
Z–1
, N + 
1
H
1
[As hydrogen has no neutrons but protons only]
?E = (M
Z–1, N
+ N
H
– M
Z,N
)c
2
?
9. E
2
N = E
Z,N–1
+ 
1
0
n .
Energy released = (Initial Mass of nucleus – Final mass of nucleus)c
2
= (M
Z.N–1
+ M
0
– M
ZN
)c
2
.
10. P
32
? S
32
+ 
0 0
0 1
v ? ?
Energy of antineutrino and ?-particle 
= (31.974 – 31.972)u = 0.002 u = 0.002 ? 931 = 1.862 MeV = 1.86.
11. In ? P + e
–
We know : Half life = 0.6931 / ? (Where ? = decay constant).
Or ? = 0.6931 / 14 ?60 = 8.25 ? 10
–4
S [As half life = 14 min = 14 ? 60 sec].
Energy = [M
n
– (M
P
+ M
e
)]u = [(M
nu
– M
pu
) – M
pu
]c
2
= [0.00189u – 511 KeV/c
2
]
= [1293159 ev/c
2
– 511000 ev/c
2
]c
2
= 782159 eV = 782 Kev. ?
The Nucleus
46.2
12.
226 4 222
58 2 26
Ra Rn ? ? ?
19 19 0 0
8 9 n 0
O F v ? ? ? ?
13 25 0 0
25 12 1 0
Al MG e v
?
? ? ?
13.
64 64
Cu Ni e v
?
? ? ?
Emission of nutrino is along with a positron emission.
a) Energy of positron = 0.650 MeV.
Energy of Nutrino = 0.650 – KE of given position = 0.650 – 0.150 = 0.5 MeV = 500 Kev.
b) Momentum of Nutrino = 
19
3
8
500 1.6 10
10
3 10
?
? ?
?
?
J = 2.67 ? 10
–22
kg m/s.
14. a)
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ?
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
40 0 40
19 1 18
K e Ar
?
? ?
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ? .
b) Q = [Mass of reactants – Mass of products]c
2
= [39.964u – 39.9626u] = [39.964u – 39.9626]uc
2
= (39.964 – 39.9626) 931 Mev = 1.3034 Mev.
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
Q = (39.9640 – 39.9624)uc
2
= 1.4890 = 1.49 Mev.
40 0 40
19 1 18
K e Ar
?
? ?
Q
value
= (39.964 – 39.9624)uc
2
.
15.
6 7
3 3
Li n Li ? ? ; 
7 8
3 3
Li r Li ? ?
8 8
3 4
Li Be e v
? ?
? ? ?
8 4 4
4 2 2
Be He He ? ?
16. ?C ? ?B + ?
+
+ v
mass of C ? = 11.014u ; mass of B ? = 11.0093u
Energy liberated = (11.014 – 11.0093)u = 29.5127 Mev.
For maximum K.E. of the positron energy of v may be assumed as 0.
? Maximum K.E. of the positron is 29.5127 Mev.
17. Mass 
238Th
= 228.028726 u ; 
224
Ra = 224.020196 u ; ? = 
4
2
He ? 4.00260u
238
Th ?
224
Ra* + ?
224
Ra* ?
224
Ra + v(217 Kev)
Now, Mass of 
224
Ra* = 224.020196 ? 931 + 0.217 Mev = 208563.0195 Mev.
KE of ? = E 
226Th
– E(
224
Ra* + ?)
= 228.028726 ? 931 – [208563.0195 + 4.00260 ? 931] = 5.30383 Mev= 5.304 Mev.
18.
12
N ?
12
C* + e
+
+ v
12
C* ?
12
C + v(4.43 Mev)
Net reaction : 
12
N ?
12
C + e
+
+ v + v(4.43 Mev)
Energy of (e
+
+ v) = N
12
– (c
12
+ v)
= 12.018613u – (12)u – 4.43 = 0.018613 u – 4.43 = 17.328 – 4.43 = 12.89 Mev.
Maximum energy of electron (assuming 0 energy for v) = 12.89 Mev.
19. a) t
1/2
= 0.693 / ? [ ? ? Decay constant]
? t
1/2
= 3820 sec = 64 min.
b) Average life = t
1/2
/ 0.693 = 92 min.
c) 0.75 = 1 e
– ?t
? In 0.75 = – ?t ? t = In 0.75 / –0.00018 = 1598.23 sec. ?
20. a) 198 grams of Ag contains ? N
0
atoms.
1 ?g of Ag contains ? N
0
/198 ? 1 ?g = 
23 6
6 10 1 10
198
?
? ? ?
atoms
Page 3


46.1
THE NUCLEUS
CHAPTER - 46
1. M = Am
p
, f = M/V, m
p
= 1.007276 u
R = R
0
A
1/3
= 1.1 ? 10
–15
A
1/3
, u = 1.6605402 ? 10
–27
kg
= 
27
3
A 1.007276 1.6605402 10
4 / 3 3.14 R
?
? ? ?
? ?
= 0.300159 ? 10
18
= 3 ? 10
17
kg/m
3
.
‘f’ in CGS = Specific gravity = 3 ? 10
14
.
2. f = 
30
13 14
17
M M 4 10 1 1
V 10 10
v f 0.6 6 2.4 10
?
? ? ? ? ? ? ?
?
V = 4/3 ?R
3
.
?
14
1
10
6
? = 4/3 ? ? R
3
? R
3
= 
14
1 3 1
10
6 4
? ? ?
?
? R
3
= 
12
1 100
10
8
? ?
?
? R = ½ ? 10
4
? 3.17 = 1.585 ? 10
4
m = 15 km.
3. Let the mass of ‘ ?’ particle be xu.
‘ ?’ particle contains 2 protons and 2 neutrons.
? Binding energy = (2 ? 1.007825 u ? 1 ? 1.00866 u – xu)C
2
= 28.2 MeV (given).
? x = 4.0016 u. ?
4. Li
7
+ p ? l + ? + E ; Li
7
= 7.016u
? = 
4
He = 4.0026u ; p = 1.007276 u
E = Li
7
+ P – 2 ? = (7.016 + 1.007276)u – (2 ? 4.0026)u = 0.018076 u.
? 0.018076 ? 931 = 16.828 = 16.83 MeV. ?
5. B = (Zm
p
+ Nm
n
– M)C
2
Z = 79 ; N = 118 ; m
p
= 1.007276u ; M = 196.96 u ; m
n
= 1.008665u
B = [(79 ? 1.007276 + 118 ? 1.008665)u – Mu]c
2
= 198.597274 ? 931 – 196.96 ? 931 = 1524.302094
so, Binding Energy per nucleon = 1524.3 / 197 = 7.737.
6. a) U
238
2
He
4
+ Th
234
E = [M
u
– (N
HC
+ M
Th
)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev.
b) E = U
238
– [Th
234
+ 2n ?
0
+ 2p ?
1
]
= {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u
= 0.024712u = 23.0068 = 23.007 MeV.
7.
223
R
a
= 223.018 u ; 
209
Pb = 208.981 u ; 
14
C = 14.003 u.
223
R
a
?
209
Pb + 
14
C
?m = mass 
223
R
a 
– mass (
209
Pb + 
14
C)
? = 223.018 – (208.981 + 14.003) = 0.034.
Energy = ?M ? u = 0.034 ? 931 = 31.65 Me. ?
8. E
Z.N.
? E
Z–1
, N + P
1
? E
Z.N.
? E
Z–1
, N + 
1
H
1
[As hydrogen has no neutrons but protons only]
?E = (M
Z–1, N
+ N
H
– M
Z,N
)c
2
?
9. E
2
N = E
Z,N–1
+ 
1
0
n .
Energy released = (Initial Mass of nucleus – Final mass of nucleus)c
2
= (M
Z.N–1
+ M
0
– M
ZN
)c
2
.
10. P
32
? S
32
+ 
0 0
0 1
v ? ?
Energy of antineutrino and ?-particle 
= (31.974 – 31.972)u = 0.002 u = 0.002 ? 931 = 1.862 MeV = 1.86.
11. In ? P + e
–
We know : Half life = 0.6931 / ? (Where ? = decay constant).
Or ? = 0.6931 / 14 ?60 = 8.25 ? 10
–4
S [As half life = 14 min = 14 ? 60 sec].
Energy = [M
n
– (M
P
+ M
e
)]u = [(M
nu
– M
pu
) – M
pu
]c
2
= [0.00189u – 511 KeV/c
2
]
= [1293159 ev/c
2
– 511000 ev/c
2
]c
2
= 782159 eV = 782 Kev. ?
The Nucleus
46.2
12.
226 4 222
58 2 26
Ra Rn ? ? ?
19 19 0 0
8 9 n 0
O F v ? ? ? ?
13 25 0 0
25 12 1 0
Al MG e v
?
? ? ?
13.
64 64
Cu Ni e v
?
? ? ?
Emission of nutrino is along with a positron emission.
a) Energy of positron = 0.650 MeV.
Energy of Nutrino = 0.650 – KE of given position = 0.650 – 0.150 = 0.5 MeV = 500 Kev.
b) Momentum of Nutrino = 
19
3
8
500 1.6 10
10
3 10
?
? ?
?
?
J = 2.67 ? 10
–22
kg m/s.
14. a)
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ?
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
40 0 40
19 1 18
K e Ar
?
? ?
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ? .
b) Q = [Mass of reactants – Mass of products]c
2
= [39.964u – 39.9626u] = [39.964u – 39.9626]uc
2
= (39.964 – 39.9626) 931 Mev = 1.3034 Mev.
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
Q = (39.9640 – 39.9624)uc
2
= 1.4890 = 1.49 Mev.
40 0 40
19 1 18
K e Ar
?
? ?
Q
value
= (39.964 – 39.9624)uc
2
.
15.
6 7
3 3
Li n Li ? ? ; 
7 8
3 3
Li r Li ? ?
8 8
3 4
Li Be e v
? ?
? ? ?
8 4 4
4 2 2
Be He He ? ?
16. ?C ? ?B + ?
+
+ v
mass of C ? = 11.014u ; mass of B ? = 11.0093u
Energy liberated = (11.014 – 11.0093)u = 29.5127 Mev.
For maximum K.E. of the positron energy of v may be assumed as 0.
? Maximum K.E. of the positron is 29.5127 Mev.
17. Mass 
238Th
= 228.028726 u ; 
224
Ra = 224.020196 u ; ? = 
4
2
He ? 4.00260u
238
Th ?
224
Ra* + ?
224
Ra* ?
224
Ra + v(217 Kev)
Now, Mass of 
224
Ra* = 224.020196 ? 931 + 0.217 Mev = 208563.0195 Mev.
KE of ? = E 
226Th
– E(
224
Ra* + ?)
= 228.028726 ? 931 – [208563.0195 + 4.00260 ? 931] = 5.30383 Mev= 5.304 Mev.
18.
12
N ?
12
C* + e
+
+ v
12
C* ?
12
C + v(4.43 Mev)
Net reaction : 
12
N ?
12
C + e
+
+ v + v(4.43 Mev)
Energy of (e
+
+ v) = N
12
– (c
12
+ v)
= 12.018613u – (12)u – 4.43 = 0.018613 u – 4.43 = 17.328 – 4.43 = 12.89 Mev.
Maximum energy of electron (assuming 0 energy for v) = 12.89 Mev.
19. a) t
1/2
= 0.693 / ? [ ? ? Decay constant]
? t
1/2
= 3820 sec = 64 min.
b) Average life = t
1/2
/ 0.693 = 92 min.
c) 0.75 = 1 e
– ?t
? In 0.75 = – ?t ? t = In 0.75 / –0.00018 = 1598.23 sec. ?
20. a) 198 grams of Ag contains ? N
0
atoms.
1 ?g of Ag contains ? N
0
/198 ? 1 ?g = 
23 6
6 10 1 10
198
?
? ? ?
atoms
The Nucleus
46.3
Activity = ?N = 
1/ 2
0.963
N
t
? = 
17
0.693 6 10
198 2.7
? ?
?
disintegrations/day.
= 
17
0.693 6 10
198 2.7 3600 24
? ?
? ? ?
disintegration/sec = 
17
10
0.693 6 10
198 2.7 36 24 3.7 10
? ?
? ? ? ? ?
curie = 0.244 Curie.
b) A = 
0
1/ 2
A 0.244
7
2t
2
2.7
?
?
= 0.0405 = 0.040 Curie. ?
21. t
1/2
= 8.0 days ; A
0
= 20 ? Cl
a) t = 4.0 days ; ? = 0.693/8
? A = A
0
e
– ?t
= 20 ? 10
–6
?
( 0.693 / 8) 4
e
? ?
= 1.41 ? 10
–5
Ci = 14 ? Ci
b) ? = 
0.693
8 24 3600 ? ?
= 1.0026 ? 10
–6
. ?
22. ? = 4.9 ? 10
–18
s
–1
a) Avg. life of 
238
U = 
18
18
1 1 1
10
4.9 4.9 10
?
?
? ? ?
? ?
sec.
= 6.47 ? 10
3
years.
b) Half life of uranium = 
18
0.693 0.693
4.9 10
?
?
? ?
= 4.5 ? 10
9
years.
c) A = 
1/ 2
0
t / t
A
2
?
1/ 2
t / t 0
A
2
A
? = 2
2
= 4.
23. A = 200, A
0
= 500, t = 50 min
A = A
0
e
– ?t
or 200 = 500 ? e
–50 ? 60 ? ?
? ? = 3.05 ? 10
–4
s.
b) t
1/2
= 
0.693 0.693
0.000305
?
?
= 2272.13 sec = 38 min. ?
24. A
0
= 4 ? 10
5
disintegration / sec
A ? = 1 ? 10
6
dis/sec ; t = 20 hours.
A ? = 
1/ 2 1/ 2
1/ 2
t / t t / t 0 0
t / t
A A
2 2 4
A ' 2
? ? ? ?
?
1/ 2
t / t = 2 ? t
1/2
= t/2 = 20 hours / 2 = 10 hours.
A ? = 
1/ 2
6
0
t / t 100 /10
A 4 10
A
2 2
?
? ? ? ? = 0.00390625 ? 10
6
= 3.9 ? 10
3
dintegrations/sec.
25. t
1/2
= 1602 Y ; Ra = 226 g/mole ; Cl = 35.5 g/mole.
1 mole RaCl
2
= 226 + 71 = 297 g
297g = 1 mole of Ra.
0.1 g = 
1
0.1
297
? mole of Ra =
23
0.1 6.023 10
297
? ?
= 0.02027 ? 10
22
? = 0.693 / t
1/2
= 1.371 ? 10
–11
.
Activity = ?N = 1.371 ? 10
–11
? 2.027 ? 10
20
= 2.779 ? 10
9
= 2.8 ? 10
9
disintegrations/second. ?
26. t
1/2
= 10 hours, A
0
= 1 ci
Activity after 9 hours = A
0
e
– ?t
= 
0.693
9
10
1 e
?
?
? = 0.5359 = 0.536 Ci.
No. of atoms left after 9
th
hour, A
9
= ?N
9
? N
9
= 
10
9
A 0.536 10 3.7 10 3600
0.693
? ? ? ?
?
?
= 28.6176 ? 10
10
?  3600 = 103.023 ? 10
13
.
Activity after 10 hours = A
0
e
– ?t
= 
0.693
9
10
1 e
?
?
? = 0.5 Ci.
No. of atoms left after 10
th
hour
A
10
= ?N
10
Page 4


46.1
THE NUCLEUS
CHAPTER - 46
1. M = Am
p
, f = M/V, m
p
= 1.007276 u
R = R
0
A
1/3
= 1.1 ? 10
–15
A
1/3
, u = 1.6605402 ? 10
–27
kg
= 
27
3
A 1.007276 1.6605402 10
4 / 3 3.14 R
?
? ? ?
? ?
= 0.300159 ? 10
18
= 3 ? 10
17
kg/m
3
.
‘f’ in CGS = Specific gravity = 3 ? 10
14
.
2. f = 
30
13 14
17
M M 4 10 1 1
V 10 10
v f 0.6 6 2.4 10
?
? ? ? ? ? ? ?
?
V = 4/3 ?R
3
.
?
14
1
10
6
? = 4/3 ? ? R
3
? R
3
= 
14
1 3 1
10
6 4
? ? ?
?
? R
3
= 
12
1 100
10
8
? ?
?
? R = ½ ? 10
4
? 3.17 = 1.585 ? 10
4
m = 15 km.
3. Let the mass of ‘ ?’ particle be xu.
‘ ?’ particle contains 2 protons and 2 neutrons.
? Binding energy = (2 ? 1.007825 u ? 1 ? 1.00866 u – xu)C
2
= 28.2 MeV (given).
? x = 4.0016 u. ?
4. Li
7
+ p ? l + ? + E ; Li
7
= 7.016u
? = 
4
He = 4.0026u ; p = 1.007276 u
E = Li
7
+ P – 2 ? = (7.016 + 1.007276)u – (2 ? 4.0026)u = 0.018076 u.
? 0.018076 ? 931 = 16.828 = 16.83 MeV. ?
5. B = (Zm
p
+ Nm
n
– M)C
2
Z = 79 ; N = 118 ; m
p
= 1.007276u ; M = 196.96 u ; m
n
= 1.008665u
B = [(79 ? 1.007276 + 118 ? 1.008665)u – Mu]c
2
= 198.597274 ? 931 – 196.96 ? 931 = 1524.302094
so, Binding Energy per nucleon = 1524.3 / 197 = 7.737.
6. a) U
238
2
He
4
+ Th
234
E = [M
u
– (N
HC
+ M
Th
)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev.
b) E = U
238
– [Th
234
+ 2n ?
0
+ 2p ?
1
]
= {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u
= 0.024712u = 23.0068 = 23.007 MeV.
7.
223
R
a
= 223.018 u ; 
209
Pb = 208.981 u ; 
14
C = 14.003 u.
223
R
a
?
209
Pb + 
14
C
?m = mass 
223
R
a 
– mass (
209
Pb + 
14
C)
? = 223.018 – (208.981 + 14.003) = 0.034.
Energy = ?M ? u = 0.034 ? 931 = 31.65 Me. ?
8. E
Z.N.
? E
Z–1
, N + P
1
? E
Z.N.
? E
Z–1
, N + 
1
H
1
[As hydrogen has no neutrons but protons only]
?E = (M
Z–1, N
+ N
H
– M
Z,N
)c
2
?
9. E
2
N = E
Z,N–1
+ 
1
0
n .
Energy released = (Initial Mass of nucleus – Final mass of nucleus)c
2
= (M
Z.N–1
+ M
0
– M
ZN
)c
2
.
10. P
32
? S
32
+ 
0 0
0 1
v ? ?
Energy of antineutrino and ?-particle 
= (31.974 – 31.972)u = 0.002 u = 0.002 ? 931 = 1.862 MeV = 1.86.
11. In ? P + e
–
We know : Half life = 0.6931 / ? (Where ? = decay constant).
Or ? = 0.6931 / 14 ?60 = 8.25 ? 10
–4
S [As half life = 14 min = 14 ? 60 sec].
Energy = [M
n
– (M
P
+ M
e
)]u = [(M
nu
– M
pu
) – M
pu
]c
2
= [0.00189u – 511 KeV/c
2
]
= [1293159 ev/c
2
– 511000 ev/c
2
]c
2
= 782159 eV = 782 Kev. ?
The Nucleus
46.2
12.
226 4 222
58 2 26
Ra Rn ? ? ?
19 19 0 0
8 9 n 0
O F v ? ? ? ?
13 25 0 0
25 12 1 0
Al MG e v
?
? ? ?
13.
64 64
Cu Ni e v
?
? ? ?
Emission of nutrino is along with a positron emission.
a) Energy of positron = 0.650 MeV.
Energy of Nutrino = 0.650 – KE of given position = 0.650 – 0.150 = 0.5 MeV = 500 Kev.
b) Momentum of Nutrino = 
19
3
8
500 1.6 10
10
3 10
?
? ?
?
?
J = 2.67 ? 10
–22
kg m/s.
14. a)
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ?
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
40 0 40
19 1 18
K e Ar
?
? ?
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ? .
b) Q = [Mass of reactants – Mass of products]c
2
= [39.964u – 39.9626u] = [39.964u – 39.9626]uc
2
= (39.964 – 39.9626) 931 Mev = 1.3034 Mev.
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
Q = (39.9640 – 39.9624)uc
2
= 1.4890 = 1.49 Mev.
40 0 40
19 1 18
K e Ar
?
? ?
Q
value
= (39.964 – 39.9624)uc
2
.
15.
6 7
3 3
Li n Li ? ? ; 
7 8
3 3
Li r Li ? ?
8 8
3 4
Li Be e v
? ?
? ? ?
8 4 4
4 2 2
Be He He ? ?
16. ?C ? ?B + ?
+
+ v
mass of C ? = 11.014u ; mass of B ? = 11.0093u
Energy liberated = (11.014 – 11.0093)u = 29.5127 Mev.
For maximum K.E. of the positron energy of v may be assumed as 0.
? Maximum K.E. of the positron is 29.5127 Mev.
17. Mass 
238Th
= 228.028726 u ; 
224
Ra = 224.020196 u ; ? = 
4
2
He ? 4.00260u
238
Th ?
224
Ra* + ?
224
Ra* ?
224
Ra + v(217 Kev)
Now, Mass of 
224
Ra* = 224.020196 ? 931 + 0.217 Mev = 208563.0195 Mev.
KE of ? = E 
226Th
– E(
224
Ra* + ?)
= 228.028726 ? 931 – [208563.0195 + 4.00260 ? 931] = 5.30383 Mev= 5.304 Mev.
18.
12
N ?
12
C* + e
+
+ v
12
C* ?
12
C + v(4.43 Mev)
Net reaction : 
12
N ?
12
C + e
+
+ v + v(4.43 Mev)
Energy of (e
+
+ v) = N
12
– (c
12
+ v)
= 12.018613u – (12)u – 4.43 = 0.018613 u – 4.43 = 17.328 – 4.43 = 12.89 Mev.
Maximum energy of electron (assuming 0 energy for v) = 12.89 Mev.
19. a) t
1/2
= 0.693 / ? [ ? ? Decay constant]
? t
1/2
= 3820 sec = 64 min.
b) Average life = t
1/2
/ 0.693 = 92 min.
c) 0.75 = 1 e
– ?t
? In 0.75 = – ?t ? t = In 0.75 / –0.00018 = 1598.23 sec. ?
20. a) 198 grams of Ag contains ? N
0
atoms.
1 ?g of Ag contains ? N
0
/198 ? 1 ?g = 
23 6
6 10 1 10
198
?
? ? ?
atoms
The Nucleus
46.3
Activity = ?N = 
1/ 2
0.963
N
t
? = 
17
0.693 6 10
198 2.7
? ?
?
disintegrations/day.
= 
17
0.693 6 10
198 2.7 3600 24
? ?
? ? ?
disintegration/sec = 
17
10
0.693 6 10
198 2.7 36 24 3.7 10
? ?
? ? ? ? ?
curie = 0.244 Curie.
b) A = 
0
1/ 2
A 0.244
7
2t
2
2.7
?
?
= 0.0405 = 0.040 Curie. ?
21. t
1/2
= 8.0 days ; A
0
= 20 ? Cl
a) t = 4.0 days ; ? = 0.693/8
? A = A
0
e
– ?t
= 20 ? 10
–6
?
( 0.693 / 8) 4
e
? ?
= 1.41 ? 10
–5
Ci = 14 ? Ci
b) ? = 
0.693
8 24 3600 ? ?
= 1.0026 ? 10
–6
. ?
22. ? = 4.9 ? 10
–18
s
–1
a) Avg. life of 
238
U = 
18
18
1 1 1
10
4.9 4.9 10
?
?
? ? ?
? ?
sec.
= 6.47 ? 10
3
years.
b) Half life of uranium = 
18
0.693 0.693
4.9 10
?
?
? ?
= 4.5 ? 10
9
years.
c) A = 
1/ 2
0
t / t
A
2
?
1/ 2
t / t 0
A
2
A
? = 2
2
= 4.
23. A = 200, A
0
= 500, t = 50 min
A = A
0
e
– ?t
or 200 = 500 ? e
–50 ? 60 ? ?
? ? = 3.05 ? 10
–4
s.
b) t
1/2
= 
0.693 0.693
0.000305
?
?
= 2272.13 sec = 38 min. ?
24. A
0
= 4 ? 10
5
disintegration / sec
A ? = 1 ? 10
6
dis/sec ; t = 20 hours.
A ? = 
1/ 2 1/ 2
1/ 2
t / t t / t 0 0
t / t
A A
2 2 4
A ' 2
? ? ? ?
?
1/ 2
t / t = 2 ? t
1/2
= t/2 = 20 hours / 2 = 10 hours.
A ? = 
1/ 2
6
0
t / t 100 /10
A 4 10
A
2 2
?
? ? ? ? = 0.00390625 ? 10
6
= 3.9 ? 10
3
dintegrations/sec.
25. t
1/2
= 1602 Y ; Ra = 226 g/mole ; Cl = 35.5 g/mole.
1 mole RaCl
2
= 226 + 71 = 297 g
297g = 1 mole of Ra.
0.1 g = 
1
0.1
297
? mole of Ra =
23
0.1 6.023 10
297
? ?
= 0.02027 ? 10
22
? = 0.693 / t
1/2
= 1.371 ? 10
–11
.
Activity = ?N = 1.371 ? 10
–11
? 2.027 ? 10
20
= 2.779 ? 10
9
= 2.8 ? 10
9
disintegrations/second. ?
26. t
1/2
= 10 hours, A
0
= 1 ci
Activity after 9 hours = A
0
e
– ?t
= 
0.693
9
10
1 e
?
?
? = 0.5359 = 0.536 Ci.
No. of atoms left after 9
th
hour, A
9
= ?N
9
? N
9
= 
10
9
A 0.536 10 3.7 10 3600
0.693
? ? ? ?
?
?
= 28.6176 ? 10
10
?  3600 = 103.023 ? 10
13
.
Activity after 10 hours = A
0
e
– ?t
= 
0.693
9
10
1 e
?
?
? = 0.5 Ci.
No. of atoms left after 10
th
hour
A
10
= ?N
10
The Nucleus
46.4
? N
10
= 
10
10
A 0.5 3.7 10 3600
0.693 /10
? ? ?
?
?
= 26.37 ? 10
10
? 3600 = 96.103 ? 10
13
.
No.of disintegrations = (103.023 – 96.103) ? 10
13
= 6.92 ? 10
13
.
27. t
1/2
= 14.3 days ; t = 30 days = 1 month
As, the selling rate is decided by the activity, hence A
0
= 800 disintegration/sec.
We know, A = A
0
e
– ?t
[ ? = 0.693/14.3]
A = 800 ? 0.233669 = 186.935 = 187 rupees. ?
28. According to the question, the emission rate of ? rays will drop to half when the ?+ decays to half of its 
original amount. And for this the sample would take 270 days.
? The required time is 270 days.
29. a) P ? n + e
+
+ v Hence it is a ?
+
decay.
b) Let the total no. of atoms be 100 N
0
.
Carbon Boron
Initially 90 N
0
10 N
0
Finally 10 N
0
90 N
0
Now, 10 N
0
= 90 N
0
e
– ?t
? 1/9 = 
0.693
t
20.3
e
?
?
[because t
1/2
= 20.3 min]
? In 
1 0.693 2.1972 20.3
t t
9 20.3 0.693
? ?
? ? ? = 64.36 = 64 min. ?
30. N = 4 ? 10
23
; t
1/2
= 12.3 years.
a) Activity = 
23
1/ 2
dN 0.693 0.693
n N 4 10
dt t 12.3
? ? ? ? ? ? dis/year.
= 7.146 ? 10
14
dis/sec.
b)
dN
dt
? 7.146 ? 10
14
No.of decays in next 10 hours = 7.146 ? 10
14
? 10 ? 36.. = 257.256 ? 10
17
= 2.57 ? 10
19
.
c) N = N
0
e
– ?t
= 4 ? 10
23
?
0.693
6.16
20.3
e
?
?
= 2.82 ? 10
23
= No.of atoms remained 
? No. of atoms disintegrated = (4 – 2.82) ? 10
23
= 1.18 ? 10
23
.
31. Counts received per cm
2
= 50000 Counts/sec.
N = N
3
o of active nucleic = 6 ? 10
16
Total counts radiated from the source = Total surface area ? 50000 counts/cm
2
= 4 ? 3.14 ? 1 ? 10
4
? 5 ? 10
4
= 6.28 ? 10
9
Counts = dN/dt
We know, 
dN
dt
? ?N
Or ? = 
9
16
6.28 10
6 10
?
?
= 1.0467 ? 10
–7
= 1.05 ? 10
–7
s
–1
. ?
32. Half life period can be a single for all the process. It is the time taken for 1/2 of the uranium to convert to 
lead.
No. of atoms of U
238
= 
23 3
6 10 2 10
238
?
? ? ?
= 
20
12
10
238
? = 0.05042 ? 10
20
No. of atoms in Pb = 
23 3
20
6 10 0.6 10 3.6
10
206 206
?
? ? ?
? ?
Initially total no. of uranium atoms = 
20
12 3.6
10
235 206
? ?
? ?
? ?
? ?
= 0.06789
N = N
0
e
– ?t
? N = N
0
1/ 2
0.693
t / t
e
?
? 0.05042 = 0.06789 
9
0.693
4.47 10
e
?
?
?
9
0.05042 0.693t
log
0.06789 4.47 10
? ? ?
?
? ?
? ? ?
? t = 1.92 ? 10
9
years. ?
1 cm
2
1 m
Page 5


46.1
THE NUCLEUS
CHAPTER - 46
1. M = Am
p
, f = M/V, m
p
= 1.007276 u
R = R
0
A
1/3
= 1.1 ? 10
–15
A
1/3
, u = 1.6605402 ? 10
–27
kg
= 
27
3
A 1.007276 1.6605402 10
4 / 3 3.14 R
?
? ? ?
? ?
= 0.300159 ? 10
18
= 3 ? 10
17
kg/m
3
.
‘f’ in CGS = Specific gravity = 3 ? 10
14
.
2. f = 
30
13 14
17
M M 4 10 1 1
V 10 10
v f 0.6 6 2.4 10
?
? ? ? ? ? ? ?
?
V = 4/3 ?R
3
.
?
14
1
10
6
? = 4/3 ? ? R
3
? R
3
= 
14
1 3 1
10
6 4
? ? ?
?
? R
3
= 
12
1 100
10
8
? ?
?
? R = ½ ? 10
4
? 3.17 = 1.585 ? 10
4
m = 15 km.
3. Let the mass of ‘ ?’ particle be xu.
‘ ?’ particle contains 2 protons and 2 neutrons.
? Binding energy = (2 ? 1.007825 u ? 1 ? 1.00866 u – xu)C
2
= 28.2 MeV (given).
? x = 4.0016 u. ?
4. Li
7
+ p ? l + ? + E ; Li
7
= 7.016u
? = 
4
He = 4.0026u ; p = 1.007276 u
E = Li
7
+ P – 2 ? = (7.016 + 1.007276)u – (2 ? 4.0026)u = 0.018076 u.
? 0.018076 ? 931 = 16.828 = 16.83 MeV. ?
5. B = (Zm
p
+ Nm
n
– M)C
2
Z = 79 ; N = 118 ; m
p
= 1.007276u ; M = 196.96 u ; m
n
= 1.008665u
B = [(79 ? 1.007276 + 118 ? 1.008665)u – Mu]c
2
= 198.597274 ? 931 – 196.96 ? 931 = 1524.302094
so, Binding Energy per nucleon = 1524.3 / 197 = 7.737.
6. a) U
238
2
He
4
+ Th
234
E = [M
u
– (N
HC
+ M
Th
)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev.
b) E = U
238
– [Th
234
+ 2n ?
0
+ 2p ?
1
]
= {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u
= 0.024712u = 23.0068 = 23.007 MeV.
7.
223
R
a
= 223.018 u ; 
209
Pb = 208.981 u ; 
14
C = 14.003 u.
223
R
a
?
209
Pb + 
14
C
?m = mass 
223
R
a 
– mass (
209
Pb + 
14
C)
? = 223.018 – (208.981 + 14.003) = 0.034.
Energy = ?M ? u = 0.034 ? 931 = 31.65 Me. ?
8. E
Z.N.
? E
Z–1
, N + P
1
? E
Z.N.
? E
Z–1
, N + 
1
H
1
[As hydrogen has no neutrons but protons only]
?E = (M
Z–1, N
+ N
H
– M
Z,N
)c
2
?
9. E
2
N = E
Z,N–1
+ 
1
0
n .
Energy released = (Initial Mass of nucleus – Final mass of nucleus)c
2
= (M
Z.N–1
+ M
0
– M
ZN
)c
2
.
10. P
32
? S
32
+ 
0 0
0 1
v ? ?
Energy of antineutrino and ?-particle 
= (31.974 – 31.972)u = 0.002 u = 0.002 ? 931 = 1.862 MeV = 1.86.
11. In ? P + e
–
We know : Half life = 0.6931 / ? (Where ? = decay constant).
Or ? = 0.6931 / 14 ?60 = 8.25 ? 10
–4
S [As half life = 14 min = 14 ? 60 sec].
Energy = [M
n
– (M
P
+ M
e
)]u = [(M
nu
– M
pu
) – M
pu
]c
2
= [0.00189u – 511 KeV/c
2
]
= [1293159 ev/c
2
– 511000 ev/c
2
]c
2
= 782159 eV = 782 Kev. ?
The Nucleus
46.2
12.
226 4 222
58 2 26
Ra Rn ? ? ?
19 19 0 0
8 9 n 0
O F v ? ? ? ?
13 25 0 0
25 12 1 0
Al MG e v
?
? ? ?
13.
64 64
Cu Ni e v
?
? ? ?
Emission of nutrino is along with a positron emission.
a) Energy of positron = 0.650 MeV.
Energy of Nutrino = 0.650 – KE of given position = 0.650 – 0.150 = 0.5 MeV = 500 Kev.
b) Momentum of Nutrino = 
19
3
8
500 1.6 10
10
3 10
?
? ?
?
?
J = 2.67 ? 10
–22
kg m/s.
14. a)
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ?
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
40 0 40
19 1 18
K e Ar
?
? ?
40 40 0 0
19 20 1 0
K Ca e v
?
? ? ? .
b) Q = [Mass of reactants – Mass of products]c
2
= [39.964u – 39.9626u] = [39.964u – 39.9626]uc
2
= (39.964 – 39.9626) 931 Mev = 1.3034 Mev.
40 40 0 0
19 18 1 0
K Ar e v
?
? ? ?
Q = (39.9640 – 39.9624)uc
2
= 1.4890 = 1.49 Mev.
40 0 40
19 1 18
K e Ar
?
? ?
Q
value
= (39.964 – 39.9624)uc
2
.
15.
6 7
3 3
Li n Li ? ? ; 
7 8
3 3
Li r Li ? ?
8 8
3 4
Li Be e v
? ?
? ? ?
8 4 4
4 2 2
Be He He ? ?
16. ?C ? ?B + ?
+
+ v
mass of C ? = 11.014u ; mass of B ? = 11.0093u
Energy liberated = (11.014 – 11.0093)u = 29.5127 Mev.
For maximum K.E. of the positron energy of v may be assumed as 0.
? Maximum K.E. of the positron is 29.5127 Mev.
17. Mass 
238Th
= 228.028726 u ; 
224
Ra = 224.020196 u ; ? = 
4
2
He ? 4.00260u
238
Th ?
224
Ra* + ?
224
Ra* ?
224
Ra + v(217 Kev)
Now, Mass of 
224
Ra* = 224.020196 ? 931 + 0.217 Mev = 208563.0195 Mev.
KE of ? = E 
226Th
– E(
224
Ra* + ?)
= 228.028726 ? 931 – [208563.0195 + 4.00260 ? 931] = 5.30383 Mev= 5.304 Mev.
18.
12
N ?
12
C* + e
+
+ v
12
C* ?
12
C + v(4.43 Mev)
Net reaction : 
12
N ?
12
C + e
+
+ v + v(4.43 Mev)
Energy of (e
+
+ v) = N
12
– (c
12
+ v)
= 12.018613u – (12)u – 4.43 = 0.018613 u – 4.43 = 17.328 – 4.43 = 12.89 Mev.
Maximum energy of electron (assuming 0 energy for v) = 12.89 Mev.
19. a) t
1/2
= 0.693 / ? [ ? ? Decay constant]
? t
1/2
= 3820 sec = 64 min.
b) Average life = t
1/2
/ 0.693 = 92 min.
c) 0.75 = 1 e
– ?t
? In 0.75 = – ?t ? t = In 0.75 / –0.00018 = 1598.23 sec. ?
20. a) 198 grams of Ag contains ? N
0
atoms.
1 ?g of Ag contains ? N
0
/198 ? 1 ?g = 
23 6
6 10 1 10
198
?
? ? ?
atoms
The Nucleus
46.3
Activity = ?N = 
1/ 2
0.963
N
t
? = 
17
0.693 6 10
198 2.7
? ?
?
disintegrations/day.
= 
17
0.693 6 10
198 2.7 3600 24
? ?
? ? ?
disintegration/sec = 
17
10
0.693 6 10
198 2.7 36 24 3.7 10
? ?
? ? ? ? ?
curie = 0.244 Curie.
b) A = 
0
1/ 2
A 0.244
7
2t
2
2.7
?
?
= 0.0405 = 0.040 Curie. ?
21. t
1/2
= 8.0 days ; A
0
= 20 ? Cl
a) t = 4.0 days ; ? = 0.693/8
? A = A
0
e
– ?t
= 20 ? 10
–6
?
( 0.693 / 8) 4
e
? ?
= 1.41 ? 10
–5
Ci = 14 ? Ci
b) ? = 
0.693
8 24 3600 ? ?
= 1.0026 ? 10
–6
. ?
22. ? = 4.9 ? 10
–18
s
–1
a) Avg. life of 
238
U = 
18
18
1 1 1
10
4.9 4.9 10
?
?
? ? ?
? ?
sec.
= 6.47 ? 10
3
years.
b) Half life of uranium = 
18
0.693 0.693
4.9 10
?
?
? ?
= 4.5 ? 10
9
years.
c) A = 
1/ 2
0
t / t
A
2
?
1/ 2
t / t 0
A
2
A
? = 2
2
= 4.
23. A = 200, A
0
= 500, t = 50 min
A = A
0
e
– ?t
or 200 = 500 ? e
–50 ? 60 ? ?
? ? = 3.05 ? 10
–4
s.
b) t
1/2
= 
0.693 0.693
0.000305
?
?
= 2272.13 sec = 38 min. ?
24. A
0
= 4 ? 10
5
disintegration / sec
A ? = 1 ? 10
6
dis/sec ; t = 20 hours.
A ? = 
1/ 2 1/ 2
1/ 2
t / t t / t 0 0
t / t
A A
2 2 4
A ' 2
? ? ? ?
?
1/ 2
t / t = 2 ? t
1/2
= t/2 = 20 hours / 2 = 10 hours.
A ? = 
1/ 2
6
0
t / t 100 /10
A 4 10
A
2 2
?
? ? ? ? = 0.00390625 ? 10
6
= 3.9 ? 10
3
dintegrations/sec.
25. t
1/2
= 1602 Y ; Ra = 226 g/mole ; Cl = 35.5 g/mole.
1 mole RaCl
2
= 226 + 71 = 297 g
297g = 1 mole of Ra.
0.1 g = 
1
0.1
297
? mole of Ra =
23
0.1 6.023 10
297
? ?
= 0.02027 ? 10
22
? = 0.693 / t
1/2
= 1.371 ? 10
–11
.
Activity = ?N = 1.371 ? 10
–11
? 2.027 ? 10
20
= 2.779 ? 10
9
= 2.8 ? 10
9
disintegrations/second. ?
26. t
1/2
= 10 hours, A
0
= 1 ci
Activity after 9 hours = A
0
e
– ?t
= 
0.693
9
10
1 e
?
?
? = 0.5359 = 0.536 Ci.
No. of atoms left after 9
th
hour, A
9
= ?N
9
? N
9
= 
10
9
A 0.536 10 3.7 10 3600
0.693
? ? ? ?
?
?
= 28.6176 ? 10
10
?  3600 = 103.023 ? 10
13
.
Activity after 10 hours = A
0
e
– ?t
= 
0.693
9
10
1 e
?
?
? = 0.5 Ci.
No. of atoms left after 10
th
hour
A
10
= ?N
10
The Nucleus
46.4
? N
10
= 
10
10
A 0.5 3.7 10 3600
0.693 /10
? ? ?
?
?
= 26.37 ? 10
10
? 3600 = 96.103 ? 10
13
.
No.of disintegrations = (103.023 – 96.103) ? 10
13
= 6.92 ? 10
13
.
27. t
1/2
= 14.3 days ; t = 30 days = 1 month
As, the selling rate is decided by the activity, hence A
0
= 800 disintegration/sec.
We know, A = A
0
e
– ?t
[ ? = 0.693/14.3]
A = 800 ? 0.233669 = 186.935 = 187 rupees. ?
28. According to the question, the emission rate of ? rays will drop to half when the ?+ decays to half of its 
original amount. And for this the sample would take 270 days.
? The required time is 270 days.
29. a) P ? n + e
+
+ v Hence it is a ?
+
decay.
b) Let the total no. of atoms be 100 N
0
.
Carbon Boron
Initially 90 N
0
10 N
0
Finally 10 N
0
90 N
0
Now, 10 N
0
= 90 N
0
e
– ?t
? 1/9 = 
0.693
t
20.3
e
?
?
[because t
1/2
= 20.3 min]
? In 
1 0.693 2.1972 20.3
t t
9 20.3 0.693
? ?
? ? ? = 64.36 = 64 min. ?
30. N = 4 ? 10
23
; t
1/2
= 12.3 years.
a) Activity = 
23
1/ 2
dN 0.693 0.693
n N 4 10
dt t 12.3
? ? ? ? ? ? dis/year.
= 7.146 ? 10
14
dis/sec.
b)
dN
dt
? 7.146 ? 10
14
No.of decays in next 10 hours = 7.146 ? 10
14
? 10 ? 36.. = 257.256 ? 10
17
= 2.57 ? 10
19
.
c) N = N
0
e
– ?t
= 4 ? 10
23
?
0.693
6.16
20.3
e
?
?
= 2.82 ? 10
23
= No.of atoms remained 
? No. of atoms disintegrated = (4 – 2.82) ? 10
23
= 1.18 ? 10
23
.
31. Counts received per cm
2
= 50000 Counts/sec.
N = N
3
o of active nucleic = 6 ? 10
16
Total counts radiated from the source = Total surface area ? 50000 counts/cm
2
= 4 ? 3.14 ? 1 ? 10
4
? 5 ? 10
4
= 6.28 ? 10
9
Counts = dN/dt
We know, 
dN
dt
? ?N
Or ? = 
9
16
6.28 10
6 10
?
?
= 1.0467 ? 10
–7
= 1.05 ? 10
–7
s
–1
. ?
32. Half life period can be a single for all the process. It is the time taken for 1/2 of the uranium to convert to 
lead.
No. of atoms of U
238
= 
23 3
6 10 2 10
238
?
? ? ?
= 
20
12
10
238
? = 0.05042 ? 10
20
No. of atoms in Pb = 
23 3
20
6 10 0.6 10 3.6
10
206 206
?
? ? ?
? ?
Initially total no. of uranium atoms = 
20
12 3.6
10
235 206
? ?
? ?
? ?
? ?
= 0.06789
N = N
0
e
– ?t
? N = N
0
1/ 2
0.693
t / t
e
?
? 0.05042 = 0.06789 
9
0.693
4.47 10
e
?
?
?
9
0.05042 0.693t
log
0.06789 4.47 10
? ? ?
?
? ?
? ? ?
? t = 1.92 ? 10
9
years. ?
1 cm
2
1 m
The Nucleus
46.5
33. A
0
= 15.3 ; A = 12.3 ; t
1/2
= 5730 year
? = 
1
1/ 2
0.6931 0.6931
yr
T 5730
?
?
Let the time passed be t, 
We know A = 
t
0
0.6931
A e t 12.3
5730
? ?
? ? ? = 15.3 ? e.
? t = 1804.3 years. ?
34. The activity when the bottle was manufactured = A
0
Activity after 8 years = 
0.693
8
12.5
0
A e
?
?
Let the time of the mountaineering = t years from the present
A = 
0.693
t
12.5
0
A e
?
?
; A = Activity of the bottle found on the mountain.
A = (Activity of the bottle manufactured 8 years before) ? 1.5%
?
0.693
12.5
0
A e
?
= 
0.693
8
12.5
0
A e 0.015
?
?
?
?
0.693 0.693 8
t In[0.015]
12.5 12.5
? ? ?
? ?
? 0.05544 t = 0.44352 + 4.1997  ? t = 83.75 years.
35. a) Here we should take R
0
at time is t
0
= 30 ? 10
9
s
–1
i) In(R
0
/R
1
) = 
9
9
30 10
In
30 10
? ?
?
? ?
? ?
?
? ?
= 0
ii) In(R
0
/R
2
) = 
9
9
30 10
In
16 10
? ?
?
? ?
? ?
?
? ?
= 0.63
iii) In(R
0
/R
3
) = 
9
9
30 10
In
8 10
? ?
?
? ?
? ?
?
? ?
= 1.35
iv) In(R
0
/R
4
) = 
9
9
30 10
In
3.8 10
? ?
?
? ?
? ?
?
? ?
= 2.06
v) In(R
0
/R
5
) = 
9
9
30 10
In
2 10
? ?
?
? ?
? ?
?
? ?
= 2.7
b) ? The decay constant ? = 0.028 min
–1
c) ? The half life period = t
1/2
.
? t
1/2
= 
0.693 0.693
0.028
?
?
= 25 min. ?
36. Given : Half life period t
1/2
= 1.30 ? 10
9 
year , A = 160 count/s = 1.30 ? 10
9
? 365 ? 86400
? A = ?N ? 160 = 
1/ 2
0.693
N
t
? N = 
9
160 1.30 365 86400 10
0.693
? ? ? ?
= 9.5 ? 10
18
? 6.023 ? 10
23
No. of present in 40 grams.
6.023 ? 10
23
= 40 g ? 1 = 
23
40
6.023 10 ?
? 9.5 ? 10
18
present in = 
18
23
40 9.5 10
6.023 10
? ?
?
= 6.309 ? 10
–4
= 0.00063.
? The relative abundance at 40 k in natural potassium = (2 ? 0.00063 ? 100)% = 0.12%.
20
15
30
10
Time t (Minute) 
25
5
25     50     75    100
Count rate R(10
9
s
–1
)