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Consider the reaction 1^2H+1^2H=2^4He+Q....

Consider the reaction `_1^2H+_1^2H=_2^4He+Q`. Mass of the deuterium atom`=2.0141u`. Mass of helium atom `=4.0024u`. This is a nuclear………reaction in which the energy Q released is…………MeV.

A

`12`

B

`6`

C

`24`

D

`48`

Text Solution

Verified by Experts

The correct Answer is:
c
`._1H^2 + .`_1H^(2) rarr `._1He^4 + Q`
`Delta m=m(._2He^4) -2 m.(_1H^2)`
`Delta m=4.0024 -2(2.0141)`
`Delta m= -0.0258u`
Now, `Q=c^(2)Delta m`
`=(0.0258)(931.5)MeV`
`~~24 MeV`.
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