Question

During a nuclear explosion, one of the products is $^{90}Sr$ with a half-life of $28.1$ years. If $1\mu g$ of $^{90}Sr$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after $10$ years and $60$ years if it is not lost metabolically.

Answer 1

Hint: Radioactive decay is basically a process in which the unstable nucleus loses its energy in the form of radiation and that substance that contains such an unstable nucleus is known as radioactive substances. So, we need to find out the amount of $^{90}Sr$left that we will find out by manipulating a formula.

Complete step by step answer:
We know that all the radioactive decay has first order reactions. We will use a formula to find out-
$k = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$, where $k$is radioactive decay constant
${N_0}$ is initial activity
$N$ is final activity
$t$ if half life
Now, first we need to find out the half-life, which we will find out using:
$k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{0.693}}{{28.1}} = 0.0247yea{r^{ - 1}}$
i) $10years$
$k = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$, using this formula we will find out the amount left.
We have,
$t = 10years$, ${N_0} = 1microgram = 1 \times {10^{ - 6}}g$,
Now, after putting these values in the above given formula,
$10 = \dfrac{{2.303}}{{0.0247}}\log \dfrac{{1 \times {{10}^{ - 6}}}}{N}$
$\log \dfrac{{1 \times {{10}^{ - 6}}}}{N} = \dfrac{{10 \times 0.0247}}{{2.303}} = 0.1072$
$\dfrac{{1 \times {{10}^{ - 6}}}}{N} = Anti\log 0.1072$
$ = 1.1280$
Now, we will take $N$ to the other side,
$N = \dfrac{{1 \times {{10}^{ - 6}}}}{{1.1280}} = 0.7842\mu g$
So, after $10$ years, $0.7842\mu g$ will be left undecayed.
ii) $60years$
Now, we will solve for $60years$ in the same way we did for $10years$
$k = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$
We have $t = 60years$, ${N_0} = 1microgram = 1 \times {10^{ - 6}}g$
After putting these values in the formula, we get:
$60 = \dfrac{{2.303}}{{0.0247}}\log \dfrac{{1 \times {{10}^{ - 6}}}}{N}$
$\dfrac{{1 \times {{10}^{ - 6}}}}{N} = Anti\log 0.6453 = 4.400$
After taking N to the other side, we get:
$N = \dfrac{{1 \times {{10}^{ - 6}}}}{{4.400}} = 0.227\mu g$
Therefore, $0.227\mu g$ will be left undecayed after $60years$

Note:
In radioactive substances, when there is a spontaneous breakdown of its atomic nucleus then that results in the emission of radiation from that nucleus and we call that as Radioactive decay.