Question

One-half of the total activity in a particular sample is due to${}^{38}\mathrm{Cl}\left(t_{1/2}=87.2\min \right).$ The other half of the activity is due to ${}^{35}\mathrm{S}\left(t_{1/2}=37.5\right.$ days). The beta emission of${}^{35}\mathrm{S}$must be measured because this nuclide emits no gamma photons. Therefore, it is desirable to wait until the activity of the${}^{38}\mathrm{Cl}$ has decreased to a negligible level. How much time must elapse before the activity of the ${}^{38}\mathrm{Cl}$ has decreased to only $0.1\%$ of the remaining activity because of ${}^{35}\mathrm{S}$ ?

Short answer

The time required to reduce activity of ${}^{38}\mathrm{Cl}$to $0.1\%$of activity of remaining ${}^{35}\mathrm{S}$islocalid="1646204171453" $14.77\mathrm{h}$.

Step-by-step solution

Given information

State the time elapsed before the activity of ${}^{38}\mathrm{Cl}$has decreased to only 0.1% of the remaining activity because of${}^{35}\mathrm{S}$/

Explanation

Write the expression for the half-life for radioactive nucleus ${}^{38}\mathrm{Cl}$.

$t_{\frac{1}{2}}=\frac{0.693}{\lambda }\dots ..\left(\mathrm{I}\right)$

Here, the half-life is $t_{\frac{1}{2}}$and the decay constant is $\lambda$.

Write the expression for the decay law.

$A=A_o{e}^{-\lambda t}\dots \dots \text{(II)}$

Here, the activity of the radioactive element after time t is A, initial activity of radioactive element is $A_o$and time period is t.

Substitute $87.2\min$for half-life of ${}^{38}\mathrm{Cl}$in Equation (1).

$$\begin{gathered} 87.2\min =\frac{0.693}{\lambda } \\ \lambda \left({}^{38}\mathrm{Cl}\right)=\frac{0.693}{87.2\min \left(\frac{1\mathrm{h}}{6\text{min}}\right)} \\ =0.4768{\mathrm{h}}^{-1} \end{gathered}$$

Substitute 37.5 days for half-life of ${}^{35}\mathrm{S}$in Equation (I).

$37.5\text{days}=\frac{0.693}{\lambda }$

$$\begin{gathered} \lambda \left({}^{35}\mathrm{S}\right)=\frac{0.693}{37.5\text{days}\left(\frac{24\mathrm{h}}{1\text{diyy}}\right)} \\ =7.7\times {10}^{-4}{\mathrm{h}}^{-1} \end{gathered}$$

Substitute value of decay-constant for ${ }^{38} \mathrm{Cl}$ in Equation (II).

$A\left({}^{38}\mathrm{Cl}\right)=A_o\left({}^{38}\mathrm{Cl}\right)e^{-\left(0.4768{\mathrm{h}}^{-1}\right)t}\dots \dots \text{(III)}$

Substitute value of decay-constant for ${}^{35}\mathrm{S}$in Equation (II).

$A\left({}^{35}\mathrm{S}\right)=A_o\left({}^{35}\mathrm{S}\right)e^{-\left(7.7\times {10}^{-4}{\mathrm{h}}^{-1}\right)t}\dots \dots \text{(IV)}$

Divide Equation (III) by Equation (IV).

Substitute 0.001 for $\frac{A\left({}^{33}\mathrm{Cl}\right)}{A\left({}^{35}\mathrm{S}\right)},A_o\left({}^{38}\mathrm{Cl}\right)$for $A_0\left({}^{35}\mathrm{S}\right)$ in Equation V.

$$\begin{gathered} 0.001=\frac{e^{-\left(0.775s^{-1}\right)t}}{e^{-\left(7.7\times {10}^{-4}{\mathrm{h}}^{-1}\right)t}} \\ -6.908=\left(-0.4768{\mathrm{h}}^{-1}\right)t+\left(7.7\times {10}^{-4}{\mathrm{h}}^{-1}\right)t \\ t=\frac{6.908}{0.47603} \\ =14.77\mathrm{h} \end{gathered}$$