Find the binding energy per nucleon for 50Sn120. Given that mass of proton mp = 1.00783amu, mass of neutron mn = 1.00867amu and mass of (Tin) mSn = 119.9021amu. (Take 1 amu = 931 MeV)

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Find the binding energy per nucleon for ₅₀Sn¹²⁰. Given that mass of proton m_p = 1.00783amu, mass of neutron m_n = 1.00867amu and mass of (Tin) m_Sn = 119.9021amu. (Take 1 amu = 931 MeV)

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9 MeV

8 MeV

7.5 MeV

8.5 MeV

answer is D.

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Detailed Solution

Mass Defect $∆$ m
$\begin{array}{l} Δm = (ZMP + (A - Z) mn - M_{Sn}) \\ Δm = 50 \times 1.00873 + 70 \times 1.008 - 119.9021 \\ Δm = 1.096 u \\ B . E = Δm \times 931 MeV \\ = 1.096 \times 931 MeV \\ BE = 1020 \times 931 MeV \\ \frac{B . E}{Nucleon} = \frac{BE}{A} = \frac{1020.56}{120} MeV \end{array}$
= 8.5 MeV