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The binding energy per nucleon of deuter...

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`

C

`26.9 MeV`

D

`13.9 MeV`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `._1 H^2 + ._1 H^2 rarr ._2 He^4 +` energy
Binding energy of a `(._1 H^2)` deuterium nuclei
`=2 xx 1.1 = 2.2 MeV`
Total binding energy of two deuterium nuclei
`=2.2 xx 2 = 4.4 MeV`
Binding energy of a `(._2 He^4)` nuclei
`=4 xx 7 = 28 MeV`
So, energy released in fusion
`=28 - 4.4 = 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, the energy released is

    A
    19.2MeV
    B
    23.6MeV
    C
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    D
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    28.0 MeV
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    30.2 MeV

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    13.9 MeV
    B
    26.9 MeV
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