1.

Consider the following reaction : `H_(1)^(2) + H_(1)^(2) = He_(2)^(4) + Q` Mass of the deuterium atom ` = 2.0141 u` 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`

Answer» Correct Answer - 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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