Question

The half-life of radium is 1620 years and its atomic weight is 226 kilograms per kilo moles. The number of atoms that will decay from its $1\,{\text{g}}$ sample per second will be: (Avogadro’s number ${N_{\text{A}}} = 6.023 \times {10^{23}}\,{\text{atoms}}\,{\text{mo}}{{\text{l}}^{ - 1}}$ )
A. $3.61 \times {10^{10}}$
B. $3.6 \times {10^{12}}$
C. $3.11 \times {10^{15}}$
D. $31.1 \times {10^{15}}$

Answer 1

First of all, we will find the decay constant of the sample. Then we will find the number of atoms contained in the given mass of the sample. After that we will use the radioactive law to find the number of atoms that will actually decay.

Complete step by step solution:
In the given question, we are given the following data:
The half-life of a sample of radium is given as $1620$ years. The atomic weight of the element radium is $226$ kilograms per kilo moles. We are also given the Avogadro’s number $N = 6.023 \times {10^{23}}\,{\text{atoms}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.
We are asked to find out the number of atoms that will decay from a sample of $1\,{\text{g}}$ .

To begin with, this is a problem from nuclear physics, in which we have to deal with the half-life of radium and then find the number of atoms that will decay. Let us solve the problem. We know, half-life of a radioactive sample is given by the following expression:
$\lambda = \dfrac{{0.693}}{{{T_{1/2}}}}$ …… (1)
Where,
$\lambda$ indicates the decay constant.
${T_{1/2}}$ indicates the half life time of the sample of radium.

Now, we substitute the required values in the equation (1) and we get:
$\lambda = \dfrac{{0.693}}{{{T_{1/2}}}} \\\ \Rightarrow \lambda = \dfrac{{0.693}}{{1620\,{\text{y}}}} \\\ \Rightarrow \lambda = \dfrac{{0.693}}{{1620 \times 365 \times 24 \times 60 \times 60\,{\text{s}}}} \\\ \Rightarrow \lambda = 1.356 \times {10^{ - 11}}\,{{\text{s}}^{ - 1}}$
Therefore, the decay constant is found to be $1.356 \times {10^{ - 11}}\,{{\text{s}}^{ - 1}}$.

Now, we will find the number of atoms in $1\,{\text{g}}$ of the sample by simply dividing the Avogadro’s number by the atomic mass of the element, which is given by:
N = \dfrac{{6.023 \times {{10}^{23}}}}{{226}} \\\ \Rightarrow N = 2.67 \times {10^{21}}\,{\text{atoms}} \\\
Where, $N$ indicates the number of atoms in $1\,{\text{g}}$ of the sample.
Since, the half-life of the sample is much larger than the time interval as compared, we can write: $N \approx \lambda \times {N_0}$

We have, according to radioactive decay law:
$dN = \lambda \times {N_0} \times dt$ …… (2)
Where,
$dN$ indicates the number of atoms that will decay.
$\lambda$ indicates the decay constant.
${N_0}$ indicates the initial number of atoms of the sample.
$dt$ indicates the time interval.

Now, we substitute the required values in the equation (2) and we get:
dN = \lambda \times {N_0} \times dt \\\ \Rightarrow dN = 1.356 \times {10^{ - 11}} \times 2.67 \times {10^{21}} \times 1 \\\ \therefore dN = 3.6 \times {10^{10}} \\\
Hence, the number of atoms that will decay from the sample is $3.6 \times {10^{10}}$. The correct option is A.

Note: While solving most of the students tend to make mistakes by taking the half-life time in years and then putting along with one second in the radioactive law. While some of them make mistakes while converting years to seconds. First, we need to convert the years into a number of days, then to hours followed by minutes and seconds.