Qstnid -3- Obtain The Binding Energy (In Mev) Of A Nitrogen Nucleus<img Src="\etl\image\esa-cbse-xii\phy-hwh-sb2/c05/1_69ad3574.gif" Width="41" Height="29">, Given <Img Src="\etl\image\esa-cbse-xii\phy-hwh-sb2/c05/1_a3f78c2.gif" Width="55" Height="29">=14.00307 U - 13: Nuclei - Neet-xii-physics

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NEET-XII-Physics

13: Nuclei

Qstn# 3 Prvs-Qstn Next-Qstn

#3
Obtain the binding energy (in MeV) of a nitrogen nucleus, given =14.00307 u

Ans : Atomic mass of nitrogen, m = 14.00307 u

A nucleus of nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7m_H + 7m_n - m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n= 1.008665 u

∴Δm = 7 × 1.007825 + 7 × 1.008665 - 14.00307

= 7.054775 + 7.06055 - 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c²

∴Δm = 0.11236 × 931.5 MeV/c²

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

E_b = Δmc²

Where,

c = Speed of light

∴E_b= 0.11236 × 931.5

= 104.66334 MeV

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