QJEEM20Ph06e7: JEE Main 2020 Physics - Class 12 JEEM20Ph06e7 Solution: Find the Binding energy per neucleon for . Mass of proton m p = …

Question

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Find the Binding energy per neucleon for

{}_{50}^{120}Sn

. Mass of proton m_p = 1.00783 U, mass of neutron m_n = 1.00867 U and mass of tin nucleus m_Sn = 119.902199 U. (take 1U = 931 MeV)

Accepted Answer

B.E. = \Delta m{c^2}

= \Delta m \times 931

\Delta m = \left( {50 \times 1.00783} \right) + \left( {70 \times 1.00867} \right) - \left\{ {119.902199} \right\}

= \left\{ {120.9984 - 119.902199} \right\}\,U

= 1.1238\,U

BE = 1.1238\, \times 931 = 1046.2578\,MeV

BE per nucleon

\simeq

1046/120

\approx

8.5 Mev

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