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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/c²]

Asked in Re-NEET 2026 · Q-value, decay and conservation laws

• (1) 3726
• (2) 3730
• (3) 3736
• (4) 3740

Step-by-step solution

Why the other options are wrong

• (2) No single slip gives 3730 keV; it would need 1 u=932.5 MeV, while the mass defect is exactly 0.004 u.
• (3) No single slip gives 3736 keV; 0.004×931.5 MeV is 3.726 MeV exactly.
• (4) No single slip gives 3740 keV; it would need a mass defect of about 0.004015 u, and the masses are given only to 0.001 u.

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