The binding energy per nucleon of 21H and 42He are 1-1 MeV and 7-0 MeV respectively- Energy released in the process 21H-21H - 42He is

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Nuclear Fission

The binding e...

Question

The binding energy per nucleon of $_{1}^{2} H$ and $_{2}^{4} H e$ are $1.1 M e V$ and $7.0 M e V$ respectively. Energy released in the process $_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e$ is

23.6 M e V

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25.2 M e V

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16.6 M e V

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13.6 M e V

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Solution

The correct option is A $23.6 M e V$
The total binding energy of $_{1}^{2} H$ = $2 \times 1.1$ = $2.2 M e V$
There are two such atoms hence
total binding energy before reaction = $2 \times 2.2 = 4.4 M e V$
Total binding energy after reaction = $4 \times$ binding energy per nucleon of $_{2}^{4} H e$
$= 4 \times 7 = 28 M e V$
Hence, energy released = $28 - 4.4 = 23.6 M e V$

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The binding energy per nucleon of 21H and 42He are 1.1 MeV and 7.0 MeV respectively. Energy released in the process 21H+21H→42He is

The correct option is A 23.6 MeVThe total binding energy of 21H = 2×1.1 =2.2 MeV There are two such atoms hence total binding energy before reaction = 2×2.2=4.4 MeV Total binding energy after reaction =4× binding energy per nucleon of 42He =4×7=28 MeV Hence, energy released = 28−4.4=23.6 MeV

The binding energy per nucleon of $_{1}^{2} H$ and $_{2}^{4} H e$ are $1.1 M e V$ and $7.0 M e V$ respectively. Energy released in the process $_{1}^{2} H +_{1}^{2} H \to_{2}^{4} H e$ is