Question

The half-life of strontium-90 is 28 years. How long will it take a 44 mg sample to decay to a mass of 11 mg?

Answer 1

Hint: The half-life of any radioactive substance is the time period required for the substance to decay, half of its initial amount. The given substance i.e. strontium-90 is the radioactive substance whose half-life is given i.e. 28 years.

Complete answer:
Let us see into radioactivity and solve the given problem;
The half-life for the radioactive sample is the interval of time required for one half of the atomic nuclei to decay. This is given by the simple equation as;
$\operatorname{Re}maining-amount=\dfrac{Initial-amount}{{{2}^{n}}}$ where, n is the number of half-lives that passed.
Now, you have given;
Half-life of strontium-90 = 28 years
Initial mass of sample = 44 mg
Remaining mass of sample = 11 mg
Putting into above stated formula, we get,
$\begin{align}
  & \operatorname{Re}maining-amount=\dfrac{Initial-amount}{{{2}^{n}}} \\
 & 11=\dfrac{44}{{{2}^{n}}} \\
 & \therefore {{2}^{n}}=\dfrac{44}{11}=4 \\
 & \Rightarrow n=2 \\
\end{align}$
Thus, two half-lives must pass;
Time required = $2\times 28=56years$ .

Note:
Do note that to demonstrate the above stated formula, general idea was used as;
${{A}_{0}}.\dfrac{1}{2}\to $ when one half-life passes
$\dfrac{{{A}_{0}}}{2}.\dfrac{1}{2}=\dfrac{{{A}_{0}}}{4}\to $ when two half-lives pass.
$\dfrac{{{A}_{0}}}{4}.\dfrac{{{A}_{0}}}{2}=\dfrac{{{A}_{0}}}{8}\to $ when three half-lives pass, and so on…
Hence, we reached the formula;
$\operatorname{Re}maining-amount=\dfrac{Initial-amount}{{{2}^{n}}}$ .