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Q.

Calculate the binding energy per nucleon of $_{20}^{40} Ca$ .Given that mass of $_{20}^{40} Ca$ nucleus = 39.962589 u. mass of a proton = 1.007825 u ; mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.

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a

8.55 Mev

b

4.55 Mev

c

6.55 Mev

d

7.55 Mev

answer is B.

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Detailed Solution

$\begin{array}{l} Z = 20, A - Z = 20; Δm = [{Zm}_{p} + (A - z) m_{n}] - M_{x} \\ Δm = 20 [1 .007825] + 20 [1 .008665] \\ - 39 .962589 = 0 .367211 u \\ ∴ \frac{BE}{A} = \frac{Δm \times 931}{40} MeV = \frac{0 .367211 \times 931}{40} \end{array}$
= 8.546MeV

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