Question

Let $N_\beta$ be the number of $\beta$ particles emitted by $1\, gram$ of $Na^{24}$ radioactive nuclei (half life = $15\, hrs$) in $7.5$ hours, $N_\beta$ is close to (Avogadro number = $6.023 \times 10^{23}$ / g mole) :

• A. $6.2 \times 10^{21}$
• B. $7.5 \times 10^{21}$
• C. $1.25 \times 10^{22}$
• D. $1.75 \times 10^{22}$

Answer 1

Answer: (B)

$N=N_{0} e^{-\lambda t}=N_{o}\left(\frac{1}{2}\right)^{x}$
Where $x=$ Number of half lives
Here $x=\frac{7.5}{15}=\frac{1}{2} h, N= \frac{N_{o}}{\sqrt{2}}$
$\Rightarrow N=\frac{6.023 \times 10^{23}}{24 \sqrt{2}}=1.75 \times 10^{22}$