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.