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Binding Energy

If the mass o...

Question

If the mass of $p r o t o n = 1.008 a . m . u .$ and mass of $n e u t r o n = 1.009 a . m . u .$ , then binding energy per nucleon for $_{4} B e^{9}$ $(m a s s = 9.012 a m u)$ would be:

40.065 M e v

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60.44 M e V

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67.2 M e v

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6.72 M e v

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Solution

The correct option is D $6.72 M e v$
mass defect is $Δ m = 4 m_{p} + (9 - 4) m_{n} - m_{B e} = (4 \times 1.008 + 5 \times 1.009 - 9.012) a m u = 0.065 a m u$

so Binding energy of $_{4} B e^{9}$ will be $Δ m \times 931.5 M e v = .065 \times 931.5 M e v = 60.5475 M e v$
or the binding energy per nucleon will be $\frac{B E}{n u c l e o n} = \frac{60.5475}{9} = 6.7275 M e v$

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_{50}^{120} S n

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