Question

A radioactive element $X$ converts into another stable element $Y$. Half life of $X$ is $2h$. Initially, only $X$ is present. After time$t$, the ratio of atoms of $X$ and $Y$ is found to be $1:4$. Then $t$ in hours is
A. 2
B. 4
C. Between 4 and 6
D. 6

Answer 1

Hint: Half life is the time when the amount of atoms becomes half of the initial amount. Here half life of element$X$is given. From that we will find the decay constant of the element$X$ and hence find the time taken to reach the given ratio of atoms.

Formula used: $N={{N}_{{{0}^{{}}}}}{{e}^{-\lambda t}}$

Complete solution:
First of all let us check what is given in the question. It is given that the half life of element $X$ is $2h$. So from this information, we will find the decay constant of this reaction by using the formula.
$N={{N}_{{{0}^{{}}}}}{{e}^{-\lambda t}}$
At half life ${{N}_{{}}}$ will be $\dfrac{1}{2}{{N}_{0}}$
So, equation becomes, $\dfrac{1}{2}{{N}_{0}}={{N}_{0}}{{e}^{-\lambda {{t}_{\dfrac{1}{2}}}}}$
Where ${{t}_{\dfrac{1}{2}}}$ is half life which is given as$2h$.
We can cancel ${{N}_{0}}$ from both sides and and substitute ${{t}_{\dfrac{1}{2}}}$
i.e., equation becomes, $\dfrac{1}{2}={{e}^{-2\lambda }}$
Rearranging the equation, we will get ${{e}^{2\lambda }}=2$
Applying natural log on both sides to eliminate$e$, equation becomes,
$2\lambda =\ln 2$
From this, $\lambda =\dfrac{\ln 2}{2}$
$\lambda =\dfrac{0.693}{2}\approx 0.3466$
So we found $\lambda =0.3466$
Now, we will find time required to become the ratio of atoms $X:Y=1:4$ .

Here the total no of atoms present is 5(i.e.$X+Y=1+4=5$), then the remaining portion of $X$ will be $\dfrac{1}{5}$.
I.e.,${{N}_{{}}}$ becomes$\dfrac{{{N}_{0}}}{5}$.
Now, the equation becomes, $\dfrac{{{N}_{0}}}{5}={{N}_{0}}{{e}^{-\lambda t}}$
We have already found$\lambda$.it will always be a constant for a reaction.
What we need to find is $t$, so we will rearrange the equation and cancel out the ${{N}_{0}}$.
$\Rightarrow$ ${{e}^{0.3466t}}=5$
Applying natural log on both sides, equation become,

& \ln 5=0.3466t \\\ & t={{\dfrac{\ln 5}{0.3466}}^{{}}}\approx \dfrac{1.6094}{0.3466}\approx 4.644h \\\ \end{aligned}$$ So the answer is option c Note: We can solve this more easily by counting half lives. At first half life $$50\%X$$ will be converted to $$Y$$. Then in the next half life the $$50\%X$$ will become $$25\%$$ and $$Y$$ become $$75\%$$. The ratio is now $$1:3$$ and time taken is $$4h$$ (2 half lives). By the next half life the ratio will be $$12.5\%:87.5\%$$ which is more than $$1:4$$. So we can conclude the time taken will be between $$4h$$ and $$6h$$.