Question

Let $N_{\beta}$ be the no of $\beta$ -particles emitted by 1 gm of $^{24}Na$ radioactive nuclei (half-life= 15 hrs. ) in 7.5 hrs, $N_{\beta}$ is close to (Avogadro number=$6.023×10^{23} /g mole$ ),

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

Answer 1

Hint: We will have to calculate the radioactive decay constant first. Then after finding the no. of atoms in 1 gm of $^{24} Na$ ,we can easily solve this problem.

Formula used:
$N=N_0 .e^{-\lambda t}$
$\lambda =\dfrac{\ln 2}{{{t}_{\dfrac{1}{2}}}}$

Complete step-by-step solution:
Half life is given as ${{t}_{\dfrac{1}{2}}}=15hrs$ . If $\lambda$ be the radioactive decay constant, we have,

$\begin{align}
  & \lambda =\dfrac{\ln 2}{{{t}_{\dfrac{1}{2}}}} \\
 & \Rightarrow \lambda =\dfrac{\ln 2}{15}=0.0462hr{{s}^{-1}} \\
\end{align}$

Now, 24 gm $^{24}Na$ contains $6.023\times {{10}^{23}}$ number of atoms.

Or, 1 gm $^{24}Na$ contains

${{N}_{0}}=\dfrac{6.023\times {{10}^{23}}}{24}=2.51\times {{10}^{22}}$ number of atoms.
The time of decay is given to be t = 7.5 hrs.

So, putting the values of $N_0$ , $\lambda$ and t , we obtain,

$N={{N}_{0}}.{{e}^{-\lambda t}}=1.77\times {{10}^{22}}$

This is the number of atoms that has remained after 7.5 hrs of decay. Hence, the number of atoms that had been decayed is,

${{N}_{\beta }}={{N}_{0}}-N=(2.51\times {{10}^{22}})-(1.77\times {{10}^{22}})=7.4\times {{10}^{21}}$

This is the number of $\beta$ particles that has been emitted.

So, option C is the correct answer.

Additional information:
A radioactive decay of a certain amount of radioactive substance takes infinite time to come to an end. It means, the process never completes and some radioactivity is always present in the substance.

Note: While calculating $\lambda$, know that (ln 2) is the natural logarithm and not the usual log (10-base) we use. Their relation is $\ln 2=2.303\times \log 2$. Another point to remember to avoid making mistakes is that N in the formula is not the number of atoms that has decayed, but the number that has remained.