Kinetic energy of the emitted $\alpha$-particle in the $\alpha$ - decay of $\frac{226}{88}$ Ra will be =
$m\left({ }_{88}^{226} Ra\right)=226.02540 u$
$m\left({ }_2^4 He\right)=4.002603 u$
$m\left({ }_{86}^{222} Rn\right)=222.01750 u$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.85 MeV
$\Delta m=(226.02540-4.002603-222.01750) u$
$\Delta m=0.005297 u$
$Q=(0.005297 u) \times\left(931.5 \frac{MeV}{u}\right)$
= 4.9341555 MeV
Hence nearby option is (B) 4.85 MeV