Question

Let $m_{p}$ be the mass of a proton $m_{n}$ the mass of a neutron $M_{1}$ the mass of ${}_{10}^{20}$ Ne nucleus and $M_{2}$ the mass of a ${}_{20}^{40}$ Ca nucleus. then

• A. $M_{2} = M_{1}$
• B. $M_{2} > 2M_{1}$
• C. $M_{2} < 2M_{1}$
• D. $M_{1} < 10(m_{n} + m_{p})$

Answer 1

Answer: (D)

$M_{1} < 10(m_{n} + m_{p})$

Answer 2

: Due to mass defect, the rest mass of a nucleus is always less than the sum of the rest masses of its constituent nucleons.

${}_{10}^{20}N$Nucleus consists of 10 protons and 10 neutrons.

$\therefore M_{1} < 10(m_{p} + m_{n})$