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Alpha Decay

Find the bind...

Question

Find the binding energy per nucleon for lithium nucleus $_{3} {L i}^{7}$ .
Take mass of $_{3} {L i}^{7} = 7 u$ , mass of proton $= 1.007825 u$ and mass of neutron $= 1.008665 u$ .
Take $1 u = 931.5 M e V$ .

3.7 M e V

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4.9 M e V

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1.7 M e V

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7.736 M e V

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Solution

The correct option is D $7.736 M e V$
Binding energy per nucleon. $E = \frac{△ E}{A} = \frac{△ m * 931}{A}$

$△ m = (3 M_{p} + 4 M_{n}) -$ Mass of $_{3} L i^{7}$

Where,

$M_{p}$ is the mass of the proton

$M_{n}$ is the mass of the neutron

$△ m =$ $(3$ X $1.007825 + 4$ X $1.008898) - 7 = 0.059067$

$△ E = 0.059067$ X $931 = 54.991377$

$E = \frac{△ E}{A} = \frac{54.991377}{7} = 7.736 M e V$

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