Question

The mass of a Li nucleus is 0.042 u less than
the sum of the masses of all its nucleons. The
binding energy per nucleon of Li nucleus is
nearly-
(1) 46 MeV
(2) 5.6 MeV
(3) 3.9 MeV
(4) 23 MeV​

Answer 1

The binding energy per nucleon of lithium nucleus is 5.6 MeV

Explanation:

We are given:

Mass defect of lithium nucleus = 0.042 u

To calculate the binding energy of the nucleus, we use the equation:

$E=\Delta mc^2\\\\E=(0.042)\times c^2$

$E=(0.042u)\times (931.5MeV)$   (Conversion factor:  $1u=931.5MeV/c^2$  )

$E=39.123MeV$

Number of nucleons in Lithium atom = 7

To calculate the binding energy per nucleon, we divide the binding energy by the number of nucleons, we get:  

$\text{Binding energy per nucleon}=\frac{\text{Binding energy}}{\text{Nucleons}}$

$\text{Binding energy per nucleon}=\frac{39.123}{7}=5.6MeV$

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