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Q.

An alpha-particle of mass m suffers 1-dimensional elastic coolision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is $______$

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answer is 4.

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Detailed Solution

From the question,

${mv}_{0} = {mv}_{2} - {mv}_{1}$

$\frac{1}{2} m {V_{1}}^{2} = 0.36 \times \frac{1}{2} m {V_{0}}^{2}$

$v_{1} = 0.6 v_{0}$

$\frac{1}{2} M {V_{2}}^{2} = 0.64 \times \frac{1}{2} m {V_{0}}^{2}$

$V_{2} = \sqrt{\frac{m}{M}} \times 0.8 V_{0}$

${mV}_{0} = \sqrt{mM} \times 0.8 V_{0} - m \times 0.6 V_{0}$

$1.6 m = 0.8 \sqrt{mM}$

$4 m^{2} = mM \Rightarrow M = 4 m$

Hence the correct answer is 4.

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