Question

A $1{\text{ }}g$ radioactive element reduces to $\dfrac{1}{5}{\text{ }}g$ after $2$ days. After a total $4$ days, how much mass will remain?
A. $\dfrac{1}{5}{\text{ }}g$
B.$\dfrac{1}{{25}}{\text{ }}g$
C. $\dfrac{1}{{125}}\,g$
D. $\dfrac{1}{{10}}{\text{ }}g$

Answer 1

Hint:From radioactive decay law, we find out that the decay constant or the disintegration constant from the formula. After finding the decay constant we will put it into the same radioactive decay formula as mentioned below and eventually we can find out the remaining undecayed mass after any time interval.

Complete step by step answer:
From the Law of Radioactive Decay we find out that, $N = {N^0}{e^{ - \lambda t}}$ where ${N^0}$ is defined as the total amount of substance present, $N$ is defined as the amount of substance remaining after time $t$ and $\lambda $ is the disintegration constant or decay constant. In the given question,
$N = \dfrac{1}{5}{\text{ }}g$
$\Rightarrow {N^0} = 1{\text{ }}g$
And $t = 2$ days
Putting the values of $N$, ${N^0}$ and $t$ in$N = {N^0}{e^{ - \lambda t}}$ we get,
$\therefore $ $N = {N^0}{e^{ - \lambda t}}$
$ \Rightarrow \dfrac{1}{5} = 1 \times {e^{ - 2\lambda }}$

Now, by putting log we can derive the formula into simpler form,
$\Rightarrow \ln \dfrac{1}{5} = - 2\lambda \\
\Rightarrow \ln 5 = 2\lambda \\ $
Now, by cross-multiplication we can find the value of decay constant $\lambda $,
$ \Rightarrow \lambda = \dfrac{{\ln 5}}{2} - - - - - \left( 1 \right)$
We have now found out the radioactive disintegration constant or decay constant.According to the question, we have to find the amount of substance remaining after $4$ days.Let us assume that the amount of substance left after $t' = $$4$ days be$N'$.

Here, $N'$ is to be found, $t = 4$ days as mentioned in the question and we have found out the radioactive decay constant or disintegration constant in equation $\left( 1 \right)$.
From Radioactive decay law we find out,
$N' = {N^0}{e^{ - \lambda t'}}$
$\Rightarrow N' = 1 \times {e^{ - \dfrac{{\ln 5}}{2} \times 4}}{\text{ putting }}\lambda {\text{ from (1)}} \\
\Rightarrow N' = {e^{ - 2\ln 5}} \\ $
Converting $x\ln y = \ln {y^x}$ formula in logarithm we get,
$\Rightarrow N' = {e^{ - \ln 25}} \\
\Rightarrow N' = {e^{\ln \dfrac{1}{{25}}}} \\ $
Now we can find the undecayed mass,
$ \therefore N' = \dfrac{1}{{25}}$
So, we have found out that after $4$ days $\dfrac{1}{{25}}{\text{ }}g$ of radioactive elements remains.

So, the correct option is B.

Note: ${e^{\ln x}} = x$ is a formula or the students might get confused. There are also some alternative method to find out this question in integration method of radioactive decay law which is being presented here too, $\int\limits_{{N^0}}^N {\dfrac{{dN}}{N} = - \lambda \int\limits_{{t^0}}^t {dt} } $. By putting the values of the respective variables we could also easily find out such types of sum easily.