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Alpha Decay

Let mp be the...

Question

Let $m_{p}$ be the mass of a proton, $m_{n}$ the mass of a neutron, $M_{1}$ is the mass of a $_{10}^{20} Ne$ nucleus & $M_{2}$ is the mass of a $_{20}^{40} Ca$ nucleus. Then

A

M_{2} = 2 M_{1}

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B

M_{2} > 2 M_{1}

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C

M_{2} < M_{1} + 10 (m_{p} + m_{n})

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D

M_{1} < 10 (m_{n} + m_{p})

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Solution

The correct option is D $M_{1} < 10 (m_{n} + m_{p})$
Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is always less than the sum of masses of its constituent particles.

$_{10}^{20} N e$ is made up of $10$ protons and $10$ neutrons. Therefore, mass of $_{10}^{20} N e$ Nucleus
$M_{1} < 10 (m_{p} + m_{n})$

Also, heavier the nucleus, more is the mass defect.

Thus , $20 (m_{n} + m_{p}) - M_{2} > 10 (m_{p} + m_{n}) - M_{1}$

$10 (m_{p} + m_{n}) > M_{2} - M_{1}$

$M_{2} < M_{1} + 10 (m_{p} + m_{n})$

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