A) 12
B) 6
C) 24
D) 48
Correct Answer: C
Solution :
[c] \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\to }_{2}}H{{e}^{4}}+Q\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\Delta m=m{{(}_{2}}H{{e}^{4}})-2m{{(}_{1}}{{H}^{2}})\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\Delta m=4.0024-2(2.0141)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\Delta m=-0.0258\,amu\] Since, \[Q={{c}^{2}}\Delta m\] \[\Rightarrow \,\,\,\,\,\,\,\,\,Q=(0.0258)(931.5)MeV\Rightarrow Q=24\,MeV.\]You need to login to perform this action.
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