Ncert Exercises & Solutions

13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus $\left({ }_{7}^{14} \mathrm{~N}\right)$ , given mass of nitrogen nucleus, $m_\left({ }_{7}^{14} \mathrm{~N}\right)$ =14.00307 u.

Hint: The binding energy can be obtained by finding the mass defect in the nitrogen nucleus.
Step 1: Find the mass defect in the nitrogen nucleus.

The atomic mass of

${}_{7}^{14}N$ nitrogen, m = 14.00307 u

A nucleus of

${}_{7}^{14}N$ nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus,

$∆ m = 7 m_{H} + 7 m_{n} - m$

where,

Mass of a proton,

$m_{H}$ = 1.007825 u
mass of a neutron,

$m_{n}$ = 1.008665 u

Mass defect,

$∴ ∆ m = 7 \times 1.007825 + 7 \times 1.008665 - 14.00307 = 0.11236 u$

$But 1 u = 931.5 MeV / c^{2}$

$∴ ∆ m = 0.111236 \times 931.5 MeV / c^{2}$

Step 2: Find the binding energy of nitrogen nucleus.

Hence, the binding energy of the nucleus is given as:

$E_{b} = ∆ {mc}^{2}$ where, c= speed of light

$∴ E_{b} = 0.11236 \times 931.5 (\frac{MeV}{c^{2}}) c^{2} = 104.66334 MeV$

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

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