Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

Consider the following nuclear reaction

238^{238}U → 234^{234}Th + 4^4He

Take masses of 238^{238}U, 234^{234}Th and 4^4He 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^{-2}]

Show answer and solution
Sol. Q-value for the nuclear reaction is given by, Q = Δmc2^2

238^{238}U → 234^{234}Th + 4^4He has mass defect Δm = mreactants_{reactants} - mproducts_{products}

⇒ Δm = 238.050 – (234.043 + 4.003) = 0.004 u

∴ Q = 0.004 × 931.5 MeV = 3.726 MeV = 3726 keV

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

More NEET UG questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans