Consider the following nuclear reaction : ^ 238 U ^ 234 Th + ^4 He Take masses of ^ 238 U , ^ 234 Th and ^4 He as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is : [Given : 1 u =931.5 MeV c^ -2 ]

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Correct answer: B. 3726

The mass defect m for the given nuclear reaction is: m = m(^ 238 U ) - [m(^ 234 Th ) + m(^4 He )] m = 238.050 - (234.043 + 4.003) m = 238.050 - 238.046 = 0.004 u The Q value of the reaction is given by: Q = m 931.5 MeV Q = 0.004 931.5 MeV Q = 3.726 MeV Converting to keV: Q = 3.726 10^3 keV = 3726 keV Answer: 3726

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