1.

In the nuclear fission ._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(4) the masses of ._(1)H^(2) and ._(2)he^(4) are 2.014 mu and 4.003 mu respectively. The energy released/atom of helium formed is ....MeV

Answer»

16.76
26.38
13.26
23.275

Solution :`Delta m = (2 XX 2.014 - 4.003) m u = 0.023"mu"`
1 amu `~~ 936 MeV`
`E = 0.025 xx 936`
E per nucleon `= (93.6)/(4) = 23.4MeV`


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