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In the following reaction, the energy re...

In the following reaction, the energy released is
`4_(1)H^(2) to 2He^(4) + 2e^(+)` + Energy. (Given: Mass of` _(1)H^(2) = 4.031300` amu, mass of `He^(4)=4.0026603` amu, mass of 2 ,`""_(1)e^(0)= 0.001098` amu).

A

12.33 MeV

B

24.67 MeV

C

25.6 MeV

D

40.34 MeV

Text Solution

Verified by Experts

The correct Answer is:
C
`4_(1)H^(2) to 2He^(4) + 2e^(+)` + Energy
Mass defect,`Delta m = 4.031300 - (4.0026603 + 0.001098)`
= 0.02754 amu
Energy released,`E = Delta m xx 931 MeV` = 2506 MeV.
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