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 ]

Step-by-step solution

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