Question

The mean life of a hypothetical nuclide is T seconds. If initially m gram of that hypothetical nuclide is present then calculate the activity of the sample at a given time t. The molar mass of the element is M.

• A. $\frac{1}{T}(\frac{M}{m}\times N_A)e^{\frac{-t}{T}}$
• B. $\frac{1}{T}(\frac{m}{M}\times N_A)e^{\frac{-t}{T}}$
• C. $\frac{1}{T}(\frac{m}{M}\times N_A)e^{\frac{-T}{t}}$
• D. $T(\frac{m}{M}\times N_A)e^{\frac{-t}{T}}$

Answer 1

The correct option is B $\frac{1}{T}(\frac{m}{M}\times N_A)e^{\frac{-t}{T}}$

Let’s proceed exactly the way gave you the hint.
Let's list out all the variables:
Mean life;
$t_{avg}=T$
Initial mass of nuclide present = m g.
Now activity at a time "t" is
$A=A_{\circ }{e}^{-\lambda t}$
To calculate activity at a time t hence, we see from the formula that we need the following:
$A_{\circ }$ - The initial activity.
$\lambda$ - The rate constant.
Now, let’s find these one by one from the given data.
We know $A_{\circ }=\lambda N_{\circ }$.
Let's find out $\lambda$ first.
We have $t_{avg}$, hence we will try relating $t_{avg}$ to $\lambda$.
$t_{avg}=\frac{t_{1/2}}{0.693}=\frac{0.693}{\lambda \times 0.693}=\frac{1}{\lambda }$
Which gives us,
$\lambda =\frac{1}{T}$.
Now let's find $N_{\circ }$.
We have initial mass = m gram.
Now, since the molar mass is M, it implies that M gram has one mole of the atom and hence nuclei.
From this m gram will have $\frac{m}{M}$ moles of nuclei.
Now, 1 mole nuclei means $N_A$ number of nuclei, hence $\frac{m}{M}$ moles will have, $\frac{m}{M}\times N_A$ nuclei.
Hence $N_{\circ }=\frac{m}{M}\times N_A$.
Now we are ready to write activity
$A=\lambda N_{\circ }{e}^{-\lambda t}=\frac{1}{T}(\frac{m}{M}\times N_A)e^{\frac{-t}{T}}$.