Question
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
A.
$${M_2} = 2{M_1}$$
B.
$${M_1} < 10\left( {{m_p} + {m_n}} \right)$$
C.
$${M_2} > 2{M_1}$$
D.
$${M_1} = {M_2}$$
Answer :
$${M_2} = 2{M_1}$$
Solution :
$$_{10}^{20}Ne$$ contains 10 protons and 10 neutrons
$$\therefore \,\,{M_1} = 10\,{m_p} + 10{m_n}$$
$$_{20}^{40}Ca$$ contains 20 protons and 20 neutrons
$$\eqalign{
& \therefore \,\,{M_2} = 20\,{m_p} + 20\,{m_n} \cr
& \therefore \,\,{M_2} = 2{M_1} \cr} $$