Question

It takes $5.2$ minutesfor a $1.000-g$ sample of ${}^{210}\mathrm{Fr}$ to decay to $0.250\mathrm{g}$. What is the half-life of ${}^{210}\mathrm{Fr}$ ?

Short answer

The half-life of ${}^{210}\text{Fr}$ is approximately $2.492$ minutes.

Step-by-step solution

Identify given values

The given values are: - Initial mass, $N_0$ = 1.000 g - Decayed mass, $N(t)$ = 0.250 g - Decay time, $t$ = 5.2 minutes

Calculate the decay constant

We'll use the decay formula to find the decay constant: $$N(t)=N_0{e}^{-\lambda t}$$ $$\frac{N(t)}{N_0}=e^{-\lambda t}$$ $$\lambda =\frac{-\ln \left(\frac{N(t)}{N_0}\right)}{t}$$ Now substitute the given values: $$\lambda =\frac{-\ln \left(\frac{0.250}{1.000}\right)}{5.2}$$ $$\lambda \approx 0.2785{\text{min}}^{-1}$$

Calculate the half-life

Now we'll use the half-life formula and the calculated decay constant: $$t_{1/2}=\frac{\ln 2}{\lambda }$$ Substitute the decay constant: $$t_{1/2}=\frac{\ln 2}{0.2785}$$ $$t_{1/2}\approx 2.492\text{minutes}$$ So the half-life of ${}^{210}\text{Fr}$ is approximately $2.492$ minutes.