Question

The mass of a ${}_3^{7}Li$ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of ${}_3^{7}Li$ nucleus is nearly

• A. 46 MeV
• B. 5.4 MeV
• C. 3.9 MeV
• D. 23 MeV

Answer 1

Answer: (B)

5.4 MeV

$\begin{array}{cc}\Delta m& =0.042\\ & =0.042+2\times 1.66\times {10}^{-27}kg\\ \text{Binding energy}& =\Delta m{c}^2\\ \text{Binding energy per nucleon}& \\ =& \frac{\Delta m{c}^2}{7}\\ & [\mathrm{mass\; number}\text{of}li=7]\end{array}$

$\begin{array}{c}=\frac{0.042\times 1.66\times {10}^{-27}\times {(3\times {10}^8)}^2}{7}J\\ \text{To convert the energy in ev from}\\ \text{joule, we will divide it by}1.6\times {10}^{-19}\\ =\frac{0.042\times 1.66\times {10}^{-27}\times {(3\times {10}^8)}^2}{7\times 1.6\times {10}^{-19}}ev\\ =5.4\times {10}^{6}ev\\ \therefore \text{Energy}\mathrm{lost}=5.4Mev\end{array}$