NEET 2026 Physics, question 7.

Consider the following nuclear reaction :

$^{238}\mathrm{U}\to{}^{234}\mathrm{Th}+{}^4\mathrm{He}$

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