Binding Energy

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New answer posted•a year ago

Binding Energy Class 12th

The binding energy (in MeV) of a nitrogen nucleus $(_{7}^{14} N)$ is :

(Given mass of nucleus $m (_{7}^{14} N) = 14.00307 u$ , mass of proton $m_{p} = 1.00783 u$ , mass of neutron $m_{n} = 1.00867 u$ and $1 u = 931.5 M e V / c^{2}$ )

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

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

Mass defect, $(_{}_{}) (\begin{matrix} \end{matrix})$ $Δ m = (7 m_{H} + 7 m_{n}) - m (\begin{matrix} 14 \\ 7 \end{matrix}) = (7 \times 1.00783 u) + (7 \times 1.00867 u) - 14.00307 u$

$= 0.11243 u$

Binding energy $= 0.11243 \times 931.5 M e V = 104.7 M e V$

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New answer posted•a year ago

Binding Energy Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Find the binding energy per neucleon for $_{50}^{120} S n$ . Mass of proton $m_{p} = (A) 1.00783 U$ , mass of neutron $m_{n} = (A) 1.00867 U$ and mass of tin nucleus $m_{S n} = 119.902199 U$ . (take $1 U = 931 M e V$ )

4 Views|Posted a year ago

Asked by Alok Kumar Singh

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Alok Kumar Singh | Contributor-Level 10

a year ago

Sol. $\Rightarrow {\overset{?}{V}}_{1} = 2 \overset{?}{U_{c m}} - \overset{?}{U_{1}} \Rightarrow 2 [\frac{m / 2 V_{o}}{\frac{m}{2} + \frac{m}{3}}] - V_{o}$

$\Rightarrow \frac{6}{5} V_{o} - V_{o} \Rightarrow \frac{V_{0}}{5}$

$\Rightarrow λ_{o} = \frac{h c}{(\frac{m}{2} V_{o})} λ_{f} = \frac{h c}{(\frac{M}{2} \frac{V_{o}}{5})}$

$\Rightarrow Δ λ = \frac{8 h c}{{m V}_{o}}$

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Binding Energy

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The binding energy (in MeV) of a nitrogen nucleus 7 14 N is : (Given mass of nucleus m 7 14 N = 14.00307 u , mass of proton m p = 1.00783 u , mass of neutron m n = 1.00867 u and 1 u = 931.5 M e V / c 2 )

Find the binding energy per neucleon for 50 120 S n . Mass of proton m p = ( A ) 1.00783 U , mass of neutron m n = ( A ) 1.00867 U and mass of tin nucleus m S n = 119.902199 U . (take 1 U = 931 M e V )

The binding energy (in MeV) of a nitrogen nucleus $(_{7}^{14} N)$ is :

(Given mass of nucleus $m (_{7}^{14} N) = 14.00307 u$ , mass of proton $m_{p} = 1.00783 u$ , mass of neutron $m_{n} = 1.00867 u$ and $1 u = 931.5 M e V / c^{2}$ )

Find the binding energy per neucleon for $_{50}^{120} S n$ . Mass of proton $m_{p} = (A) 1.00783 U$ , mass of neutron $m_{n} = (A) 1.00867 U$ and mass of tin nucleus $m_{S n} = 119.902199 U$ . (take $1 U = 931 M e V$ )