The mass defect and binding energy per nucleon of an alpha particle are :
(Mp $=$ 1.00734 $u$ , Mn $=$ 1.00874 $u$ , M $=$ 4.0015 $u$ )

0.0303 u, 28.3 M e V

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0.1303 u, 28.3 M e V

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0.0303 u, 7.07 M e V

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0.0306 u, 7.14 M e V

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Solution

The correct option is D $0.0306 u, 7.14 M e V$
$_{2}^{4} H e$
$Δ m = [2 \times 1.00734 + 2 \times 1.00874 - 4.0015]$ $= 0.03066 u$
$B . E = 0.03066 \times 931.478 M e V$ $= 28.56 M e V$
$B . E p e r n u c l e o n = \frac{28.56}{4}$ $= 7.14 M e V$

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_{50}^{120} S n

. Mass of proton

m_{p} = 1.00783 U

, mass of neutron

m_{n} = 1.00867 U

and mass of tin nucleus

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1 U = 931 MeV)

Find the binding energy per nucleon of $_{79}^{197} A u$ if its atomic mass is 196.96 u. The atomic mass of $_{1}^{1} H$ is 1.007825 u, mass of neuton, $m_{n} = 1.008665 u, m_{p} = 1.007276 u . c^{2} = 931 \frac{M e V}{u} .$
Remember that atomic mass also contains the mass of electrons.