Kinetic energy of the emitted $\alpha$-particle in the $\alpha$ - decay of $\frac{226}{88}$ Ra will be =

$m\left({ }_{88}^{226} Ra\right)=226.02540 u$

$m\left({ }_2^4 He\right)=4.002603 u$

$m\left({ }_{86}^{222} Rn\right)=222.01750 u$

4.85 MeV

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