Question

Calculate the time interval between \[33\% \]decay and \[67\% \] decay if the half-life of a substance is 20 minutes.
A. 40 minutes
B. 20 minutes
C. 60 minutes
D. 13 minutes

Answer 1

Hint: Before we proceed with the problem let’s understand the radioactive decay. When an unstable atomic nucleus loses energy through radiation this is known as radioactive decay. This radioactive decay involves the spontaneous transformation of one element into another. This can happen by changing the number of protons in the nucleus.

Formula Used:
The equation for the radioactive decay is,
\[N = {N_0}{e^{ - \lambda t}}\]
Where,
\[N\]is the number of atoms remaining
\[{N_0}\]is the original number of atoms
\[\lambda \]is decay constant
t is the time taken

Complete Step by Step Solution:
The equation for radioactive decay is,
\[N = {N_0}{e^{ - \lambda t}}\]
We can write this as,
\[{N_1} = {N_0}{e^{ - \lambda {t_1}}}\]
Now, in \[100\% \] the \[67\% \]decay happens that is \[N = 67\% \]and \[{N_0} = 100\% \]
Then, the above equation becomes,
\[67 = 100{e^{ - \lambda t}}\]
\[0.67 = {e^{ - \lambda t}}\]
Here if we take the natural logarithm on both sides for the above equation we obtain,
\[\ln \left( {0.67} \right) = - \lambda {t_1}\]………….. (1)
\[{N_2} = {N_0}{e^{ - \lambda {t_2}}}\]
Now, in \[100\% \] the \[33\% \]decay happens that is \[N = 33\% \]and \[{N_0} = 100\% \]
Then, the above equation becomes,
\[33 = 100{e^{ - \lambda {t_2}}}\]
\[0.33 = {e^{ - \lambda {t_2}}}\]
Take natural logarithm on both sides for the above equation then, we obtain,
\[\ln \left( {0.33} \right) = - \lambda {t_2}\]………….. (2)
Subtract the equation (2) from (1) we obtain,
\[\ln \left( {0.67} \right) - \ln \left( {0.33} \right) = - \lambda {t_1} + \lambda {t_2}\]
\[\ln \left( {\dfrac{{0.67}}{{0.33}}} \right) = \lambda \left( {{t_2} - {t_1}} \right)\]
We know that, \[\lambda = \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}\] \[ \Rightarrow \lambda = \dfrac{{\ln 2}}{{20}}\]
Substitute the value in the above equation, and we obtain,
\[\ln \left( 2 \right) = \dfrac{{\ln 2}}{{20}}\left( {{t_2} - {t_1}} \right)\]
\[\left( {{t_2} - {t_1}} \right) = 20\min \]
Therefore, the time interval between\[33\% \]decay and \[67\% \]decay is 20min
Hence, Option B is the correct answer

Note: Here in the given question it is important to remember that whenever the exponential term is present in the equation. In order to find the value first, we need to eliminate the exponential term. The exponential term can be eliminated by applying a natural logarithm on both sides as shown in this problem.