Question

The activity of a radioactive sample is measured as ${N_0}$ counts per minute at $t = 0$ and $\dfrac{{{N_0}}}{e}$ counts per minute at t=5 minutes. The time (in minutes) at which the activity reduces to half its value is
A. ${\log _e}\dfrac{2}{5}$
B. $\dfrac{5}{{{{\log }_e}2}}$
C. $5{\log _{10}}2$
D. $5{\log _e}2$

Answer 1

Activity for radioactive substance for half-life period is given as$A = {A_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$
The activity of radioactive substances is defined as the rate of a number of the disintegration of the given sample per second or number of unstable atomic decay per second in the given sample. This activity is determined by counting the number of particles and pulses of electromagnetic energy by radiation detectors and electronic circuits.
Here, in the question, we need to determine the time at which the radioactive substance decays to half of its initial value. For this, we will use the formula $A = {A_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}$ and follow the arithmetic principles.

Complete step by step answer:
Activity for radioactive substance is given as $A\left( t \right) = {A_0}{e^{ - \lambda t}}$
Hence the activity for radioactive substance for half-life period will be $A = {A_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}} - - (i)$
Given at t=0 the activity of a radioactive sample is${N_0}$,
At t=5 min the activity of a radioactive sample is $\dfrac{{{N_0}}}{e}$
Hence we can say
${A_0} = {N_0}$
$A = \dfrac{{{N_0}}}{e}$
Now substitute this equation (i), we get
$\dfrac{{{N_0}}}{e} = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{5}{{{T_{\dfrac{1}{2}}}}}}}{\text{ }}[t = 5{\text{ min}}]$
This can also be written as
${e^{ - 1}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{5}{{{T_{\dfrac{1}{2}}}}}}}$
Now we apply log to both sides of the obtained equation

$$\log \left( {{e^{ - 1}}} \right) = \log \left( {{{\left( {\dfrac{1}{2}} \right)}^{\dfrac{5}{{{T_{\dfrac{1}{2}}}}}}}} \right) \\\ \- \ln \left( e \right) = \dfrac{5}{{{T_{\dfrac{1}{2}}}}}\ln \left( {\dfrac{1}{2}} \right) \\\$$

Since$\ln \left( e \right) = 1$so we can write

$$\- \ln \left( e \right) = \dfrac{5}{{{T_{\dfrac{1}{2}}}}}\ln \left( {\dfrac{1}{2}} \right) \\\ \- 1 = \dfrac{5}{{{T_{\dfrac{1}{2}}}}}\ln \left( {\dfrac{1}{2}} \right) \\\ {T_{\dfrac{1}{2}}} = - 5\ln \left( {\dfrac{1}{2}} \right) \\\$$

This can be written as

{T_{\dfrac{1}{2}}} = - 5\ln \left( {\dfrac{1}{2}} \right) \\\ {T_{\dfrac{1}{2}}} = - 5\left( {\ln 1 - \ln 2} \right) \\\ $$[Since$$\ln \dfrac{a}{b} = \ln a - \ln b$$] $${T_{\dfrac{1}{2}}} = - 5\left( {0 - \ln 2} \right)$$[Since$$\ln 1 = 0$$] Hence we obtain $${T_{\dfrac{1}{2}}} = 5\ln 2$$ Also, this can be written as $${T_{\dfrac{1}{2}}} = 5{\log _e}2$$. **So, the correct answer is “Option D”.** **Note:** It is interesting to note here that the radioactive substances decay exponentially. Radioactive substances are unstable, and the dangerous substances that produce radiation. This substance is unstable because the stronger nucleus force that holds the nucleus of the atom together is not in balance with the electric force that wants to push atoms apart, and these make radioactive substances to decay.