## Thursday, March 20, 2008

### Two EAMCET Questions from Nuclear Physics

The following MCQ which appeared in EAMCET 2005 question paper highlights the law of conservation of momentum as applied to nuclear processes:

A nucleus of mass 218 amu in free state decays to emit an α-particle. Kinetic energy of the α-particle emitted is 6.7 MeV. The recoil energy (in MeV) of the daughter nucleus is

(a) 1

(b) 0.5

(c) 0.25

(d) 0.125

The kinetic energy (K) is given by

K = p2/2m where p is the momentum and m is the mass. The recoil momentum of the daughter nucleus is equal and opposite to the momentum of the α-particle.

We have p2/(2×4 amu) = 6.7 MeV, since the mass of the α-particle is 4 amu.

[You need not convert amu into kilogram and energy into joule and waste your time. You may imagine that p has proper unit to get the energy in MeV]

In the case of the daughter nucleus (of mass 218 – 4 = 214 amu), we have

p2/(2×214 amu) = K, whetre K is the recoil energy of the daughter (in MeV).

From the above two equations we obtain K = 0.125 MeV.

The following MCQ appeared in EAMCET 2000 question paper:

In a nuclear reactor using U235 as fuel the power output is 4.8 mega watts. The number of fissions per second is (energy released per fission of U235 = 200 MeV).

(a) 1.5×1017

(b) 3×1019

(c) 1.5×1025

(d) 3×1025

This is a very simple question. The energy produced per second is 4.8×106 joule. Since 1 electron volt is 1.6×10–19 joule, 200 MeV = 200×106×1.6×10–19 joule.

Therefore, the number of fissions per second = (4.8×106)/(200×106×1.6×10–19) = 1.5×1017.

You can find all related posts on this site by clicking on the label ‘nuclear physics’ below this post.

#### 1 comment:

1. Mark Bill11:28 AM

Thanks for sharing this post with us.