Question

The half-life of radium is 1620 years and its atomic weight is 226. The number of atoms that will decay from 1g sample per second will be:
$
  {\text{A}}{\text{. 3}}{\text{.6 }} \times {\text{ 1}}{{\text{0}}^{10}} \\
  {\text{B}}{\text{. 3}}{\text{.6 }} \times {\text{ 1}}{{\text{0}}^{12}} \\
  {\text{C}}{\text{. 3}}{\text{.1 }} \times {\text{ 1}}{{\text{0}}^{15}} \\
  {\text{D}}{\text{. 31}}{\text{.1 }} \times {\text{ 1}}{{\text{0}}^{15}} \\
$

Answer 1

- Hint: Here half life is given so using ${t_{½}} = \dfrac{{0.693}}{\lambda }$ this formula we will get decay constant $\lambda $ and using decay constant and applying this formula $R = \dfrac{dN}{{dt}} = - \lambda N$ we will get required number. For the number of atoms we use the mole concept.

Formula used:
Half life: ${t_{½}} = \dfrac{{0.693}}{\lambda }$
$R = \dfrac{{dN}}{{dt}} = - \lambda N$ , where R is the activity of the sample.

Complete step-by-step answer:
In the above question, the given parameters are:
Half life of radium: ${t_{½}}$ = 1620 years
Atomic weight of radium: W = 226 kg/mol
The mass of the sample is: m = 1g
Avogadro’s number: N = 6.023 $ \times $ $10^{23}$ atoms/mol
The radioactive decay law states that the probability per unit time that a nucleus will decay is a constant which is independent of time and this constant is denoted by λ.
In radioactive decay the number of radioactive atoms decaying per unit time is proportional to the total number of radioactive atoms at the present time and therefore radioactive decay follows first order kinetics.
The measure of time for the decay of a sample by its half amount is known as the half-life and it is a characteristic constant for decaying samples.
 The relationship between the half-life, ${t_{½}}$, and the decay constant λ is given by the formula,
${t_{½}} = \dfrac{{0.693}}{\lambda }$
So, the decay constant is,
λ = $\dfrac{0.693}{{t_{½}}}$
= $\dfrac{0.693}{{1620×365×24×3600}}$
=1.356×${10^{-11}}$${s^{-1}}$

Number of atoms of radium in 1 gram is given by,
n= N/W
n=$\dfrac{6.023×10^{23}}{{226}}$
=2.67×${10^{21}}$ atoms

By the formula of radioactive decay law,
The number of atom decayed will be,
R= $\dfrac{{\text{dN}}}{{dt}}$​ = −λN
=1.356×${10^{-11}}$ × 2.67×10$^{21}$
=3.6×10$^{10}$ atoms/s
Therefore, the number of atoms decayed per second is 3.6×10$^{10}$ atoms/s.

Note: In this question it is important to know about the concepts of first order kinetics and decay laws. This question is a simple example of radioactivity. We should know radioactivity is an example of first order kinetics. And this type of questions will be easily solved if we have remembered $R = \dfrac{dN}{{dt}} = - \lambda N$
${t_{½}} = \dfrac{0.693}{{\lambda }}$ these formulae.