Let m_p be the mass of a proton,m_n that of a | vedclass

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(A) The mass of a nucleus is given by $M = Z m_p + (A-Z) m_n - 	ext{binding energy}/c^2$. For $_{10}^{20}Ne$,$Z=10$ and $A=20$,so $M_1 \approx 10 m_p

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$M_2 = 2M_1$

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(A) The mass of a nucleus is given by $M = Z m_p + (A-Z) m_n - 	ext{binding energy}/c^2$. For $_{10}^{20}Ne$,$Z=10$ and $A=20$,so $M_1 \approx 10 m_p + 10 m_n - 	ext{binding energy}_1/c^2$. For $_{20}^{40}Ca$,$Z=20$ and $A=40$,so $M_2 \approx 20 m_p + 20 m_n - 	ext{binding energy}_2/c^2$. Since the binding energy per nucleon is higher for $_{20}^{40}Ca$ than for $_{10}^{20}Ne$,the mass defect per nucleon is higher for Calcium. Therefore,$M_2 However,in the context of simplified nuclear models where binding energy is neglected,$M_1 = 10(m_p + m_n)$ and $M_2 = 20(m_p + m_n)$,which leads to $M_2 = 2 M_1$.

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

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