Find the binding energy per nucleon for ${}_{50}^{120} \mathrm{Sn}$. Mass of proton $\mathrm{m}_{\mathrm{p}}=1.00783 \mathrm{U}$, mass of neutron $m_{n}=1.00867 \mathrm{U}$ and mass of tin nucleus $m_{\mathrm{Sn}}=119.902199 \mathrm{U}$. (take $1 \mathrm{U}=931 \mathrm{MeV}$)

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Modern Physics

Find the binding energy per nucleon 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

1 U = 931 MeV

)

(A)8.5 MeV✓ Correct
(B)9.0 MeV
(C)7.5 MeV
(D)8.0 MeV

Answer: (A)

Solution

The tin nucleus

_{50}^{120} Sn

contains

50

protons and

70

neutrons, so the mass defect is

Δ m = (50 \times 1.00783 + 70 \times 1.00867) - 119.902199 = 1.0962 U

. Multiplying by

931 MeV U^{- 1}

gives a total binding energy of about

1020.6 MeV

, and dividing by the

120

nucleons yields a binding energy per nucleon of

\frac{1020.6}{120} \approx 8.5 MeV

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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