Question

The half-life of ${}^{198}$ Au is 2.7 days. Find the number of active nuclei present in a sample containing $1.414\times {10}^{10}$ nuclei after a time that is half of its half-life.

• A. $\frac{1}{2}\times 1.414\times {10}^{10}$
  
• B. ${10}^{10}$
  
• C. $2\times {10}^{10}$
  
• D. $\frac{1}{4}\times 1.414\times {10}^5$

Answer 1

The correct option is B ${10}^{10}$


Okay , let's do this!!!
First piece of information is that the half life is 2.7 days
I know $t_{1/2}=\frac{ln2}{\lambda }$
So this means $\lambda =\frac{ln2}{t_{1/2}}$
Moving on
Next piece of info is $N_0=1.414\times {10}^{10}$
And the question is asking Number of active nuclei present after a time that is half of half-life i.e. $\frac{1}{2}{t}_{1/2}$
Looks scary.
Let's put the formula and see what all we know in it.
$N=N_0{e}^{-\lambda t}$
$N=1.414\times {10}^{10}x{e}^{-}\frac{ln2}{t_{1/2}}\times \frac{t_{1/2}}{2}$ $N=1.414\times {10}^{10}x{e}^{-}\frac{ln2}{2}$
$N=1.414\times {10}^{10}\times 2^{-}\frac{1}{2}$ (That's how we do in log. remember)
$N={10}^{10}$
So the number of active nuclei left after $1\times {10}^{10}t=\frac{1}{2}{t}_{1/2}$