Consider the following nuclear reaction $^{238}U\rightarrow{}^{234}Th+{}^{4}He$ 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]

Question

Dr. Abubacker · Accepted Answer

[Solution from source] Mass defect = 238.050 − (234.043 + 4.003) = 0.004 u. Q = 0.004 × 931.5 MeV = 3.726 MeV = 3726 keV. Answer (2).

Q. Consider the following nuclear reaction $^{238}U\rightarrow{}^{234}Th+{}^{4}He$ 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]

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?