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The binding energy per nucleon of deuter...

The binding energy per nucleon of deuterium and helium nuclei are 1.1 MeV and 7.0 MeV respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

A

23.6 MeV

B

2.2 MeV

C

28.0 MeV

D

30.2 MeV

Text Solution

Verified by Experts

The correct Answer is:
A
`._1H^2 + ._1H^2 to ._2He^4 + DeltaE`
The binding energy per nucleon of a deuterium =1.1 MeV
`therefore` total binding energy = 2 x 1.1 = 2.2 MeV
The binding energy per nucleon of a helium nuclei =7 MeV
`therefore` Total binding energy = 4 x 7 =28 MeV
Hence, energy released `DeltaE`=(28-2 x 2.2 )=23.6 MeV
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Knowledge Check

  • The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV . If two deuterium nuclei fuse to form helium atom, the energy released is.

    A
    `19.2 MeV`
    B
    `23.6 MeV`
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    D
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    D
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