Chapter 30, Problem 47P

To determine

The activity of a pure $6.7\mu g$ sample of ${}_{15}{}^{32}P$ ($T_{1/2}=1.23\times {10}^{6}s$)?

Answer to Problem 47P

Solution:

The activity of a pure $6.7\mu g$ sample of ${}_{15}{}^{32}P$$=7.1\times {10}^{10}$ decays/s

Explanation of Solution

Formula used:

The number of nuclei remaining after a time t, $N=N_0{e}^{-\lambda t}$, $N_0$ is the number of nuclei present after time t = 0 and $\lambda$ is the decay constant.

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:

We have initial decay rate $=\lambda N_0=2.4\times {10}^5{s}^{-1}$

We have Decay constant, $\lambda =\frac{\ln 2}{1.23\times {10}^6}$$s^{-1}$

We also have $N_0=\frac{6.7\times {10}^{-6}\times 6.02\times {10}^{23}}{32}$

⇨ Activity $=\lambda N_0=\frac{\ln 2}{1.23\times {10}^6}\times \frac{6.7\times {10}^{-6}\times 6.02\times {10}^{23}}{32}$

$=7.1\times {10}^{10}$ decays/s