The alpha decay of polonium‑210 is given by ${}^{210}\text{Po}{\to }^{206}\text{Pb}+alpha.$ The atomic masses are $m{(}^{210}\text{Po})=209.98785\text{u}$, $m{(}^{206}\text{Pb})=205.9725\text{u}$, and $m(\alpha )=4.002603\text{u}$. Using $1\text{u}=931.5\text{MeV}/c^2$, calculate the $Q$‑value of the decay. Assume the neutrino carries negligible kinetic energy.

Step-by-step solution

1. Write the masses of parent, daughter, and alpha.
2. Compute the total mass of the products: $205.9725+4.002603=209.975103\text{u}$.
3. Find the mass defect: $209.98785-209.975103=0.012747\text{u}$.
4. Convert to energy: $0.012747\times 931.5=11.876\text{MeV}$.
5. Round to one decimal: $11.9\text{MeV}$.
6. Answer: $11.9\text{MeV}$.

Concept tested: $Q$‑value of alpha decay

Published 7 October 2026 · How this solution was prepared and checked