Consider the nuclear reaction ²³⁸U→²³⁴Th+⁴He. Taking the masses of ²³⁸U, ²³⁴Th and ⁴He as 238.050 u, 234.043 u and 4.003 u respectively, the Q-value of the reaction (in keV) is: [1 u =931.5 MeV c⁻²]

Solution & Explanation

Principle: the Q-value of a decay equals the mass lost (reactant mass minus product masses) converted to energy via E=Δm·c², with 1 u → 931.5 MeV. Step 1 — Mass difference. Reactant: ²³⁸U = 238.050 u. Products: ²³⁴Th + ⁴He = 234.043 + 4.003 = 238.046 u. Δm = 238.050 - 238.046 = 0.004 u. Step 2 — Convert to energy. Q = Δm × 931.5 MeV = 0.004 × 931.5 = 3.726 MeV. Step 3 — Convert to keV. Q = 3.726 MeV = 3726 keV. This matches option A (3726 keV). Note: Q is positive, confirming the decay is energetically allowed (spontaneous α-decay). The closeness of the options (3726/3730/3736/3740) means you must keep all decimals in the masses — rounding early would push you to a wrong choice.