Question

The half-life of radium is $1620\, yr$ and its atomic weight is $226\, kg$ per kilomol. The number of atoms that will decay from its $1\, g$ sample per second will be (Avogadro's number $N=6.023 \times 10^{23}$ atoms/ $mol$ )

• A. $3.61 \times 10^{10}$
• B. $3.61 \times 10^{12}$
• C. $3.11 \times 10^{15}$
• D. $31.1 \times 10^{15}$

Answer 1

Answer: (A)

$3.61 \times 10^{10}$

Answer 2

From Rutherford Soddy law, the number of atoms that will decay is
$\frac{d N}{d t}=\lambda N$ ...(i)
Also, $\lambda=\frac{0.693}{T_{1 / 2}}$ ...(ii)
where $T_{1 / 2}$ is half-life of radioactive sample.
Given, $T_{1 / 2}=1620\, yr$
$=1620 \times 365 \times 24 \times 60 \times 60$
$\therefore \lambda=\frac{0.693}{1620 \times 365 \times 24 \times 60 \times 60}$
and $N=\frac{6.023 \times 10^{23}}{226}$
Putting these values of $N$ and $\lambda$ in E (i), we get
$\frac{d N}{d t}=\frac{0.693 \times 6.023 \times 10^{23}}{1620 \times 365 \times 24 \times 60 \times 60 \times 226}$
$\frac{d N}{d t}=3.61 \times 10^{10}$