Question

The half-life of ${}_{238}{}^{92}\mathrm{U}$ undergoing α-decay is $4.5\times {10}^{-9}$. What is the true activity initial of $1\mathrm{gm}$ of the sample?

Answer 1

Step 1: Given data:

The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay.

The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

The half-life of ${}_{238}{}^{92}\mathrm{U}$ is given as ;

$$\begin{gathered} {\mathrm{T}}_{\frac{1}{2}}=4.5\times {10}^9 \\ =4.5\times {10}^9\times 3.156\times {10}^7\mathrm{s} \end{gathered}$$

Hence, the given data refers to the fact that the radioactive isotope ${}_{238}{}^{92}\mathrm{U}$ will take $4.5\times {10}^9\times 3.156\times {10}^7\mathrm{s}$ time for the amount of the substance to be reduced or decay to half of the initial amount.

We know Avogadro's number ${\mathrm{N}}_{\mathrm{A}}=6.023\times {10}^{23}$

Step 2: Calculating the number of atoms in 1g of Uranium:

The number of atoms contained in 1g of any substance is given by:

$\mathrm{N}=\frac{N_0}{\mathrm{molecular}\mathrm{mass}}$

$\because$Molecular mass of Uranium= $238\mathrm{gm}$

Also, Avogadro's number ${\mathrm{N}}_o=6.023\times {10}^{23}$

Hence, the number of atoms in$1\mathrm{g}$of uranium can be calculated as:

$$\begin{gathered} \mathrm{N}=\frac{{\mathrm{N}}_0}{238} \\ =\frac{6.023\times {10}^{23}}{238} \end{gathered}$$

Step 3: Formula for true activity:

The total decay rate R of a sample of one or more radionuclides is called the true activity of that sample.

The general formula for calculating the true activity of that sample can be given as:
True activity,${\mathrm{A}}_{\mathrm{o}}={\mathrm{\lambda N}}_{\mathrm{o}}$

Here, $\mathrm{\lambda }$ is denoted as the radioactivity constant, which is given by:

​​$\mathrm{\lambda }=\frac{0.693}{{\mathrm{T}}_{{\displaystyle \frac{1}{2}}}}$​

Step 4: Calculation of true activity of $1\mathrm{gm}$ of the sample:

By substituting the given data in the formula given for true activity of the given radioactive sample, true activity can be calculated as:

$$\begin{gathered} \mathrm{True}\mathrm{activity}, \\ =\mathrm{\lambda N} \\ =\frac{0.693}{\mathrm{T}\frac{1}{2}}\times \mathrm{N} \\ =\frac{0.693}{4.5\times {10}^9\times 3.15\times {10}^7}\times \frac{6.023\times {10}^{23}}{238} \\ =1.237\times {10}^4\mathrm{d}/\mathrm{s} \end{gathered}$$

Hence, the activity of the sample is $1.237\times {10}^4\mathrm{d}/\mathrm{s}$