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Question

A nucleus X initially at rest, undergoes alpha decay according to the equation,

ZX232 90YA+α

What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle.

A
228232
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B
9092
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C
228232
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D
12
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Solution

The correct option is A 228232
The given equation can be represented as,

ZX232 90Y228+ 2He4(αparticle)

As nucleus X is initially at rest,

According to the conservation of momentum,

0=PY+Pα

The momentum of nucleus Y and α particle will be same.

We know that, K.E=P22m

K.E1m

K.EαK.EY=mYmα

Here, mA

K.EαK.EY=2284

K.EαK.ETotal=K.EαK.Eα+K.EY

K.EαK.ETotal=228232

Hence, option (D) is correct.
Why this Question?

This question gives and idea of how to apply law of conservation of linear momentum in the nuclear emission.


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