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,
\[\begin{align}
  & \ln 5=0.3466t \\
 & t={{\dfrac{\ln 5}{0.3466}}^{{}}}\approx \dfrac{1.6094}{0.3466}\approx 4.644h \\
\end{align}\]
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 \[6h\].