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Binding Energy

The binding e...

Question

The binding energy per nucleon of $_{2} H e^{4}$ is $7.075$ MeV. What is its total binding energy?

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Solution

Binding energy per nucleon $= 7.075 M e V$
No. of nucleon $= 4$
So, Total Binding Energy $= 7.075 \times 4 = 28.3 M e V$

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_{2}^{4} H e

28.57

M e V

_{8} O^{16}

127 M e V

_{3}

^{7}

_{1} H^{2}

1.112 M e V

_{2} H e^{4}

7.047 M e V

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + Q

Q

_{1} H^{2} = 2.01478 u

_{2} H e^{4} = 4.00338 u

28

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