Question

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 1

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`.