Question

The radioactive isotope ${}^{137}Cs$ has a half life period of $30$ years. Starting with $1mg$ of ${}^{137}Cs$, how much would remain after $120$ years?

Answer 1

At this time, we have $1.0mg$ of ${}^{137}Cs$; after $30$ years, we shall have one half of the original, or $0.50mg$; after $60$ years, we shall have $0.25mg$; after $90$ years, we shall have $0.125mg$ and , finally after $120$ years, we shall have $0.0625mg$.
After $30$ years $-$ $0.50mg$
$60$ years $-$ $0.25mg$
$90$ years $-$ $0.125mg$
$120$ years $-$ $0.0625mg$
Alternative solution: Total time $=120$ years
We know that, total time $=n\times t_{1/2}$
So, $120=n\times 30$
$n=4$
Thus, the quantity of the isotope left after four half life periods $={\left(\frac{1}{2}\right)}^4{N}_0={\left(\frac{1}{2}\right)}^4\times 1$
$=\frac{1}{6}=0.0625mg$