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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` |
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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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