422
7 Nuclear Physics – I
Using λ 238 = 1/τ 238 = 1/6.52 × 10
9
, λ 235 = 1/τ 235 = 1/1.02 × 10
9
And solving (3) we get the age of the earth as 5.26 × 10
9 years
7.99 Number of Th atoms in 2 μg is
N = N 0 W/A = 6.02 × 10
23
× 2 × 10
−6
/224 = 5.375 × 10
15
λ = 0.693/(3.64 × 86, 400) = 2.2 × 10
−6
Activity A = |dN /d λ| = N λ = (5.375 × 10
15 )(2.2 × 10
−6 ) = 1.183 ×
10
10
/s
= (1.188 × 10
10
/3.7 × 10
10 ) ci = 0.319 ci
7.100 1 rad = 100 ergs g
−1
Energy received = (40)(100)(20 × 10
3 )/10
7 J
= 8 J
= 8/4.18 = 1.91 Cal
7.101 Let V cm
3 be the volume of blood, Initial activity is 16,000/V per minute
per cm
3 . 0.8 =
16,000
V
exp(−0.693 × 30/15)
=
4,123
V
∴ V = 5000 cm
3
7.3.10 Alpha-Decay
7.102 λ 1 = 10
21 exp(−2π zZ 1 e
2
/ v 1 )
λ 2 = 10
21 exp(−2 πzZ 2 e
2
/v 2 )
Given λ 1 = λ 2
Z 1 /
√
E 1 = Z 2 /
√
E 2
E 2 = E 1 Z 2
2
/Z 1
2
= 5.3(80/82)
2
= 5.04 MeV
Thus energy from the second nuclei is 5.04 MeV.
7.103 The energy required to force an α-particle(classically) into a nucleus of
charge Ze is equal to the potential energy at the barrier height and is given by
E = zZe
2
/4 π ε 0 R.
This is barely possible when α particle and the Uranium nucleus will just
touch each other and the kinetic energy of the bombarding particle is entirely
converted into potential energy.
R = R 1 + R 2 = r (A
1/3
1 + A 2
1/3 )
= 1.2(4
1/3
+ 238
1/3 ) = 9.34 fm
E = 1.44z Z/R = 1.44 × 2 × 92/9.34 = 28.37 MeV
7.104 Geiger–Nuttal law is
log λ = k log x + c
where k and c are constants, λ is the decay constant and x is the range
log(0.693/T ) = k log x + c
k log x + log T = log 0.693 − c = c 1
(1)
7 Nuclear Physics – I
Using λ 238 = 1/τ 238 = 1/6.52 × 10
9
, λ 235 = 1/τ 235 = 1/1.02 × 10
9
And solving (3) we get the age of the earth as 5.26 × 10
9 years
7.99 Number of Th atoms in 2 μg is
N = N 0 W/A = 6.02 × 10
23
× 2 × 10
−6
/224 = 5.375 × 10
15
λ = 0.693/(3.64 × 86, 400) = 2.2 × 10
−6
Activity A = |dN /d λ| = N λ = (5.375 × 10
15 )(2.2 × 10
−6 ) = 1.183 ×
10
10
/s
= (1.188 × 10
10
/3.7 × 10
10 ) ci = 0.319 ci
7.100 1 rad = 100 ergs g
−1
Energy received = (40)(100)(20 × 10
3 )/10
7 J
= 8 J
= 8/4.18 = 1.91 Cal
7.101 Let V cm
3 be the volume of blood, Initial activity is 16,000/V per minute
per cm
3 . 0.8 =
16,000
V
exp(−0.693 × 30/15)
=
4,123
V
∴ V = 5000 cm
3
7.3.10 Alpha-Decay
7.102 λ 1 = 10
21 exp(−2π zZ 1 e
2
/ v 1 )
λ 2 = 10
21 exp(−2 πzZ 2 e
2
/v 2 )
Given λ 1 = λ 2
Z 1 /
√
E 1 = Z 2 /
√
E 2
E 2 = E 1 Z 2
2
/Z 1
2
= 5.3(80/82)
2
= 5.04 MeV
Thus energy from the second nuclei is 5.04 MeV.
7.103 The energy required to force an α-particle(classically) into a nucleus of
charge Ze is equal to the potential energy at the barrier height and is given by
E = zZe
2
/4 π ε 0 R.
This is barely possible when α particle and the Uranium nucleus will just
touch each other and the kinetic energy of the bombarding particle is entirely
converted into potential energy.
R = R 1 + R 2 = r (A
1/3
1 + A 2
1/3 )
= 1.2(4
1/3
+ 238
1/3 ) = 9.34 fm
E = 1.44z Z/R = 1.44 × 2 × 92/9.34 = 28.37 MeV
7.104 Geiger–Nuttal law is
log λ = k log x + c
where k and c are constants, λ is the decay constant and x is the range
log(0.693/T ) = k log x + c
k log x + log T = log 0.693 − c = c 1
(1)