Question

A radioactive isotope of mercury${}^{\mathbf{197}}\mathbf{Hg}$, decays to gold${}^{\mathbf{197}}\mathbf{Au}$, with a disintegration constant of$\mathbf{0}\mathbf{.}\mathbf{0108}{\mathbf{h}}^{\mathbf{-}\mathbf{1}}$. (a) Calculate the half-life of the${}^{\mathbf{197}}\mathbf{Hg}$. What fraction of a sample will remain at the end of (b) three half-lives and (c)10days?

Short answer

a) The half-life of the isotope is 64.2h.

b) The fraction of sample that will remain at the end of three half-lives is 0.125.

c) The fraction of sample that will remain at the end of 10 days is 0.0749.

Step-by-step solution

The given data

Disintegration constant,$\mathrm{\lambda }=0.0108{\mathrm{h}}^{-1}$

Understanding the concept of decay constant and half-life

Half-life is the time that takes for the number of radioactive nuclei to decrease to half of their initial value. Its relation with the decay constant shows that the population decays exponentially at a rate that depends on the decay constant. The radioactive decay constant or the disintegration constant represents the fraction of radioactive atoms that disintegrates in a unit of time.

Formulae:

The disintegration constant is as follows:

$\mathrm{\lambda }=\frac{\mathrm{In}2}{{\mathrm{T}}_{1/2}}$ ….. (i)

Here,$T_{\frac{1}{2}}$is the half-life of the substance.

The undecayed sample remaining after a given time as follows:

$N=N_0{e}^{-\lambda t}$ ….. (ii)

a) Calculate the half-life of the isotope

Using the given data in equation (i), get the half-life of the isotope as follows:

$$\begin{gathered} T_{1/2}=\frac{\mathrm{In}2}{\lambda } \\ =\frac{\mathrm{In}2}{0.0108{\mathrm{h}}^{-1}} \\ =64.2\mathrm{h} \end{gathered}$$

Hence, the value of the half-life is 64.2h.

b) Calculate the fraction of sample that will remain after three half-lives

Using equation (i) in equation (ii) and the given data $t=3T_{1/2}$, get the undecayed sample that will remain at the end of three half-lives as follows:

$$\begin{gathered} N=N_0{e}^{\frac{\mathrm{In}2}{T_{{\displaystyle \frac{1}{2}}}}\left(3T_{\frac{1}{2}}\right)} \\ =e^{-3\mathrm{In}2} \end{gathered}$$

Solve further as:

$\frac{N}{N_0}=0.125$

In each half-life, the number of undecayed nuclei is reduced by half. At the end of one half-life $N=\frac{N_0}{2}$, at the end of two half-lives, $N=\frac{N_0}{4}$and at the end of three half-lives is given as:

$$\begin{gathered} N=\frac{N_0}{8} \\ =0.125N_0 \end{gathered}$$

Hence, the value of the fraction of sample is 0.125.

c) Calculate the fraction of sample that will remain after 10 days

Using the given data in equation (ii) for t = 10y or 240d, get the undecayed sample that will remain at the end of three half-lives as follows:

$$\begin{gathered} \frac{N}{N_0}=e^{-\left(0.0108{\mathrm{h}}^{-1}\right)\left(240d\right)} \\ =e^{-2.592} \\ =0.0749 \end{gathered}$$

Hence, the value of the fraction is 0.0749.