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Energy released in the nuclear fusion re...

Energy released in the nuclear fusion reaction is :` ""_(1)^(2)H + ""_(1)^(3)H rarr ""_(2)^(4)He + ""_(0)^(1)n`
Atomic mass of `""_(1)^(2)H=2.014 , ""_(1)^(3)H=3.016 ""_(2)^(4)He=4.303, ""_(0)^(1)n=1.009` (all in a.m.u.)

A

16.60 MeC

B

500 J

C

`4xx10^(67)` Kcal

D

8.30 eV

Text Solution

Verified by Experts

The correct Answer is:
A
`Deltam=(2.014 + 3.016)-(1.009+4.003)`
=5.030-5.012 = 0.018 amu
1 amu =931 MeV
`therefore E=0.018xx931.5` MeV=16.77MeV
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