Question

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

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

Answer 1

Answer: (C), (D)

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.
$_{10}^{20} Ne$ is made up of 10 protons plus 10 neutrons. Therefore,
mass of $_{10}^{20} Ne$ nucleus, $M_1 <10(m_p+m_n)$.
Also, heavier the nucleus, more is the mass defect.
Thus, $20(m_n+m_p)-M_2 >10(m_p+m_n)-M_1$
or $10(m_p+m_n) >M_2-M_1$
or $ M_2 < M_1+10(m_p+m_n)$
Now since $ M_1<10(m_p+m_n)$
$\therefore $ $ M_2 < 2M_1$