Consider the following nuclear reaction 238U 234Th + 4He Take masses of 238U, 234Th and 4He 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]
- 3740
- 3726
- 3730
- 3736
Answer
(2) 3726
Solution
The mass defect of the reaction is:
m = m( 238U) - [m( 234Th) + m(4He)]
m = 238.050 - (234.043 + 4.003) = 238.050 - 238.046 = 0.004 u
The Q value is:
Q = m 931.5 MeV = 0.004 931.5 MeV = 3.726 MeV = 3726 keV
Answer: (2)