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Question

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
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B
0.5
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C
0.25
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D
0.125
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Solution

The correct option is A 0.125

Conservation of total momentum of system implies:

P1(Momentum of alpha particle) + P2 (Momentum of daughter nucleus) =0 (Initial momentum)

Thus, P1=P2

M= mass of daughter nucleus = 214 amu

m= mass of alpha particle = 4 amu

E is the energy of daughter nucleus in MeV

Now , 6.7MeV=P212m, for alpha particle

E=P222M, for daughter nucleus

Dividing the above two equations, we get

E6.7=mM (as P1=P2 or P21=P22)

Therefore, E=6.7×4214=0.125


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