Question

The mass of deuteron $\left(_{1} H^{2}\right)$ nucleus is $2.014103u.$ If the masses of proton and neutron are $1.007275u$ and $1.007275u$ respectively, calculate the binding energy per nucleon.

• A. $2.2MeV$
• B. $1.1MeV$
• C. $0.5MeV$
• D. $0.25MeV$

Answer 1

Answer: (B)

Here, $m_{p}=1.007275m:m_{n}=1.008667u$ and mass of $\_{1}^{}H_{}^{2}$ nucleus, $m_{N}\left(H H^{2}\right)=2.0159403u.$ The deuteron nucleus contains one proton and one neutron. Therefore, mass of nucleons constituting deuteron, $m_{p}+m_{n}=1.007275+$
$1.008665=2.015940u.$
Therefore, mass defect, $\Delta m=\left(m_{p} + m_{n}\right)-m_{N}\left(\_{1}^{}H_{}^{2}\right)$
$=2.01540-2.013553=0.002387u$
Packing fraction, $\frac{\Delta m}{A}=\frac{0 . 002387}{2}=0.0011935u.$