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* = *p*^{2}/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 *p*^{2}/(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

*p*^{2}/(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 U ^{235} as fuel the power output is 4.8 mega watts. The number of fissions per second is (energy released per fission of U^{235} = 200 MeV).**

**(a) 1.5****×10 ^{17}**

**(b) 3****×10 ^{19}**

**(c) 1.5****×10 ^{25}**

**(d) 3****×10 ^{25}**

This is a very simple question. The energy produced per second is 4.8×10^{6} joule. Since 1 electron volt is 1.6×10^{–19} joule, 200 MeV = 200×10^{6}×1.6×10^{–19} joule.

Therefore, the number of fissions per second = (4.8×10^{6})/(200×10^{6}×1.6×10^{–19}) = 1.5×10^{17.}

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

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