1000 Solved Problems in Modern Physics

476 8 Nuclear Physics – II

The actual radius will be shorter byd= 0. 71 λtrwhered=extrapolated

distance.

8.81 Letnfissions take place per second.
Energy released per second= 200 nMeV=(200n)(1. 6 × 10 −^13 )J
200 × 1. 6 × 10 −^13 n= 40 × 106
Or n= 1. 25 × 1018 /s
Number of atoms in 1.0 g of^235 U=

6. 02 × 1023

235

= 2. 562 × 1021

Mass of uranium consumed per second=

1. 25 × 1018

2. 562 × 1021

= 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 nMeV s−^1
=(200n)(1. 6 × 10 −^13 )J-s−^1
=power= 2 × 107 W
Therefore,n= 6. 25 × 1017 s−^1
In 1 g there areN 0 /A= 6. 02 × 1023 / 235 = 2. 56 × 1021 atoms of^235 U.
Therefore consumption rate of^235 U will be
6. 25 × 1017
2. 56 × 1021

= 2. 44 × 10 −^4 gs−^1

8.83 (a) Let n fissions occur per second. Then energy available will be 200 n MeV
or 200n× 1. 6 × 10 −^13 J. Allowing for 5% wastage and 30% efficiency, net
power used isP= 200 n× 1. 6 × 10 −^13 × 0. 95 × 0. 3 = 9. 12 × 10 −^12 nW.
Required energy in 1 s that is power =

50 × 109 × 103 × 3 , 600

3. 15 × 107

=

5. 714 × 109

Equating the two powers

9. 12 × 10 −^12 n= 5. 714 × 109

n= 6. 26 × 1020 /s

Number of atoms in 1 g of^235 U=

6. 02 × 1023

235

= 2. 56 × 1021

Mass consumed per second=

6. 26 × 1020

2. 56 × 1021

= 0 .244 g

Mass consumed in 1 year= 0. 244 × 3. 15 × 107 g= 7. 7 × 106 g= 7 .7 tons

(b) Volume,V=

M

ρ

=

108

19000

= 526 .3m^3

Power density (Power per unit volume)

P/V=

100 × 106

526. 3

= 1. 9 × 105 W/m^3