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13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus ( ${}_{7}^{14}{N)}$ , given

m( ${}_{7}^{14}{N)}$ =14.00307 u

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Payal Gupta | Contributor-Level 10

3 months ago

13.3 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, Δm = 7 $m_{p}$ + 7 $m_{n}$ - m, where

Mass of a proton, $m_{p}$ = 1.007825 u

Mass of a neutron, $m_{n}$ = 1.008665 u

Therefore, Δm = 7 $\times$ 1.007825+ 7 $\times$ 1.008665 – 14.00307 = 0.11236 u

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

Δm = 104.66334 MeV/ $c^{2}$

The binding energy of the nucleus, $E_{b}$ = Δm $c^{2}$ , where c = speed of light

$E_{b} =$ (104.66334/ $c^{2}$ ) $\times$ $c^{2}$ = 104.66334 MeV

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13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus ( ${}_{7}^{14}{N)}$ , given

m( ${}_{7}^{14}{N)}$ =14.00307 u

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13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus ( ${}_{7}^{14}{N)}$ , given

m( ${}_{7}^{14}{N)}$ =14.00307 u