Question

The binding energy of ${}_{17}{CI}^{35}$ nucleus is $298MeV.$ Find its atomic mass. The mass of hydrogen atom $\left({{}_{1}H}^1\right)$ is $1.008143amu$ and that of a neutron is $1.008986amu.$ Given $1amu=931MeV.$

• A. $14.8$
• B. $29.8$
• C. $34.0$
• D. $34.8$

Answer 1

The correct option is D $34.8$

$\begin{array}{c}\text{Let mass defect be}\Delta m\\ \left[\Delta m\right]931=B\cdot E\\ \left[\Delta m\right]931=298\\ \Delta m=0.32amu\end{array}$

$\begin{array}{c}\text{Atomic weight,}M\\ \begin{array}{cc}M=& (n_p\times m_p+n_n\times m_n)-\left(\Delta m\right)\\ M& (17\times 1.008143+18\times 1.0089)-0.32\\ M& =(35.2974-0.32)amu\\ M& =34.8779\text{amu}\end{array}\end{array}$