Let m _p be the mass of a proton, m _n the mass of a neutron, M₁ the mass of a ${ }_{10}^{20}\mathrm{Ne}$ nucleus and M₂  the mass of a ${ }_{20}^{40}\mathrm{Ca}$ nucleus. Then find the wrong option(s) .

(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 i.e., protons and neutrons. $\begin{array}{r}20\\ 10\end{array}\mathrm{Ne}$ is made up of 10 protons and 10 neutrons. $\therefore {\text{ Mass of }}_{10}^{20}\text{ Ne nucleus }$ ${\mathrm{M}}_1<10\left({\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{n}}\right)$ Heavier the nucleus, more is the mass defect 
$$\begin{gathered} 20\left(m_n+m_p\right)-M_2>10\left(m_p+m_n\right)-M_1 \\ \text{ or }10\left(m_n+m_p\right)>M_2-M_1 \\ \text{ or }{\mathrm{M}}_2<{\mathrm{M}}_1+10\left({\mathrm{m}}_{\mathrm{p}}+{\mathrm{m}}_{\mathrm{n}}\right) \\ \text{ Now, since }{\mathrm{M}}_1<10\left({\mathrm{m}}_{\mathrm{n}}+{\mathrm{m}}_{\mathrm{p}}\right) \\ \therefore {\mathrm{M}}_2<2{\mathrm{M}}_1 \end{gathered}$$