476
8 Nuclear Physics – II
The actual radius will be shorter by d = 0.71 λ tr where d = extrapolated
distance.
8.81 Let n fissions take place per second.
Energy released per second = 200 n MeV = (200 n) (1.6 × 10
−13 ) J
200 × 1.6 × 10
−13 n = 40 × 10
6
Or
n = 1.25 × 10
18
/s
Number of atoms in 1.0 g of
235 U =
6.02 × 10
23
235
= 2.562 × 10
21
Mass of uranium consumed per second =
1.25 × 10
18
2.562 × 10 21 = 4.88 × 10
−4 g
Mass consumed in 1 day = (4.88 × 10
−4 )(86, 400) = 42.16 g
8.82 Let n be the number of fissions occurring per second in the nuclear reactor.
Energy released = 200 n MeV s
−1
= (200 n)(1.6 × 10
−13 )J-s
−1
= power = 2 × 10
7 W
Therefore, n = 6.25 × 10
17 s
−1
In 1 g there are N 0 /A = 6.02 × 10
23
/235 = 2.56 × 10
21 atoms of
235 U.
Therefore consumption rate of
235 U will be
6.25 × 10
17
2.56 × 10 21 = 2.44 × 10
−4 g s
−1
8.83 (a) Let n fissions occur per second. Then energy available will be 200 n MeV
or 200 n × 1.6 × 10
−13 J. Allowing for 5% wastage and 30% efficiency, net
power used is P = 200 n × 1.6 × 10
−13
× 0.95 × 0.3 = 9.12 × 10
−12 n W.
Required energy in 1 s that is power =
50 × 10
9
× 10
3
× 3, 600
3.15 × 10 7
=
5.714 × 10
9
Equating the two powers
9.12 × 10
−12 n = 5.714 × 10
9
n = 6.26 × 10
20
/s
Number of atoms in 1 g of
235 U =
6.02 × 10
23
235
= 2.56 × 10
21
Mass consumed per second =
6.26 × 10
20
2.56 × 10 21 = 0.244 g
Mass consumed in 1 year = 0.244 × 3.15 × 10
7 g = 7.7 × 10
6 g = 7.7 tons
(b) Volume, V =
M
ρ
=
10
8
19000
= 526.3 m
3
Power density (Power per unit volume)
P/V =
100 × 10
6
526.3
= 1.9 × 10
5 W/m
3