Problem 52

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Problem 52

Find the energy (in MeV) released when $\alpha$ decay converts radium ²²⁶₈₈Ra (atomic mass = 226.02540 u) into radon ²²⁶₈₆Rn (atomic mass = 222.01757 u).

SOLUTION: Use Energy Conservation. By `` $\alpha$ decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get

m = 4.00260 u + 222.01757 u = 226.02017 u .

The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is

$\begin{displaymath} \Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,\end{displaymath}$

which is equivalent to an energy change of

$\begin{displaymath} \Delta E = (0.00523 u) \left ( \frac{931.5 MeV}{1 u} \right )\end{displaymath}$

$\begin{displaymath} \Delta E = 4.87 MeV\end{displaymath}$

Scott Lanning
4/23/1998