Calculate the binding energy of a nitrogen nu | vedclass

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Question

(D) The nitrogen nucleus $\left[{ }_7^{14} Night]$ contains $Z = 7$ protons and $N = (14 - 7) = 7$ neutrons.Mass of $7$ protons $= 7 	imes 1.00783

Vedclass · Accepted Answer

$104.7$

Vedclass · Answer

(D) The nitrogen nucleus $\left[{ }_7^{14} Night]$ contains $Z = 7$ protons and $N = (14 - 7) = 7$ neutrons.Mass of $7$ protons $= 7 	imes 1.00783 \ u = 7.05481 \ u$.Mass of $7$ neutrons $= 7 	imes 1.00867 \ u = 7.06069 \ u$.Total mass of nucleons $= 7.05481 \ u + 7.06069 \ u = 14.11550 \ u$.Mass defect $\Delta m = (	ext{Total mass of nucleons}) - (	ext{Mass of nucleus})$.$\Delta m = 14.11550 \ u - 14.00307 \ u = 0.11243 \ u$.Binding energy $BE = \Delta m 	imes 931.5 \ MeV/u$.$BE = 0.11243 	imes 931.5 \approx 104.73 \ MeV$.Rounding to the nearest option,the binding energy is $104.7 \ MeV$.

Calculate the binding energy of a nitrogen nucleus $\left[{ }_7^{14} N\right]$,given that the mass of the nucleus $m\left[{ }_7^{14} N\right] = 14.00307 \ u$. (Take mass of proton $m_p = 1.00783 \ u$ and mass of neutron $m_n = 1.00867 \ u$) (in $MeV$)

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