Question

Let $m_p$ be the mass of proton, $m_n$ be the mass of neutron, $M_1$ be the mass of ${}_{10}{\text{Ne}}^{20}$ nucleus and $M_2$ be the mass of a ${}_{20}{\text{Ca}}^{40}$ nucleus. Then

• A. $M_2=2M_1$
• B. $M_2>2M_1$
• C. $M_2>20[m_n+m_p]$
• D. $M_1<10(m_n+m_p)$

Answer 1

The correct option is D $M_1<10(m_n+m_p)$

The true mass of the nucleus is always less than the sum of masses of the nucleons.

Here, $M_1-$ true mass of the ${}_{10}{\text{Ne}}^{20}$ nucleus.

Also, $10(m_n+m_p)-$ sum of masses of the nucleons of ${}_{10}{\text{Ne}}^{20}$.

So, $M_1<10(m_n+m_p)$ and $M_2<20(m_n+m_p)$

Also, from the binding energy per nucleon versus mass number curve, for lighter nuclei, the binding per nucleon increases with mass number.

Thus, the mass defect also increases with mass number for lighter nuclei.

So, $M_2<2M_1$

Hence, option $\left(D\right)$ is correct.