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 ....MeVA. 16.76B. 26.38C. 13.26D. 23.275

In the nuclear fission `._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(4)` the

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 ....MeVA. 16.76B. 26.38C. 13.26D. 23.275
Answer» Correct Answer - D `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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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 ....MeVA. 16.76B. 26.38C. 13.26D. 23.275

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