Let {m_p} be the mass of a proton,{m_n} the m | vedclass

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(D) The mass of a nucleus is always less than the sum of the masses of its constituent nucleons due to the mass defect,which accounts for the binding

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Both $(a)$ and $(c)$

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(D) The mass of a nucleus is always less than the sum of the masses of its constituent nucleons due to the mass defect,which accounts for the binding energy of the nucleus.For the $_{10}^{20}Ne$ nucleus,which consists of $10$ protons and $10$ neutrons,the mass ${M_1}$ is given by ${M_1} For the $_{20}^{40}Ca$ nucleus,which consists of $20$ protons and $20$ neutrons,the binding energy per nucleon is higher than that of $_{10}^{20}Ne$. Since the binding energy per nucleon increases with the mass number for lighter nuclei,the mass defect per nucleon is greater for $_{20}^{40}Ca$ than for $_{10}^{20}Ne$.Therefore,the mass of the $_{20}^{40}Ca$ nucleus ${M_2}$ is less than twice the mass of the $_{10}^{20}Ne$ nucleus ${M_1}$,i.e.,${M_2} Thus,both $(a)$ and $(c)$ are correct.

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

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