Chapter 30, Problem 54P

To determine

The relative decay rates for decay of ${}_{84}{}^{218}Po$ and ${}_{84}{}^{214}Po$ using the given figure.

Answer to Problem 54P

Solution:

Relative decay $=8.6\times {10}^{-7}$

Explanation of Solution

Formula used:

Decay constant, $\lambda =\frac{\ln 2}{T_{1/2}}$, $T_{1/2}$ is the half life

Activity $=\lambda N$, where $\lambda$ is the decay constant and N is the number of nuclei remaining after a time t

Calculation:

The relative decay is given by $\frac{{(\lambda N)}_{{}_{84}{}^{218}Po}}{{(\lambda N)}_{{}_{84}{}^{214}Po}}$, in this case we can assume $N_{{}_{84}{}^{218}Po}=N_{{}_{84}{}^{214}Po}$

So relative decay = $\frac{{\lambda }_{{}_{84}{}^{218}Po}}{{\lambda }_{{}_{84}{}^{214}Po}}=\frac{{(T_{1/2})}_{{}_{84}{}^{214}Po}}{{(T_{1/2})}_{{}_{84}{}^{218}Po}}$

From figure we have ${(T_{1/2})}_{{}_{84}{}^{214}Po}=1.6\times {10}^{-4}s$ and ${(T_{1/2})}_{{}_{84}{}^{218}Po}=3.1\min$

⇨ Relative decay = $\frac{1.6\times {10}^{-4}}{3.1\times 60}=8.6\times {10}^{-7}$