Question

The count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is
A. 120 min
B. 80 min
C. 30 min
D. 20 min

Answer 1

Hint: The law of radioactive decay helps you calculate the number of atoms present at the given time approximately. The half-life is also an important factor to calculate the time taken for the decay of the given sample.

Formula used:
The law of radioactive decay,
\[N = {N_0}{e^{ - \lambda t}}\]
Where, \[N\] = the amount of atoms present after a time period
\[{N_0}\] = the amount of atoms at t = 0
\[\lambda \] = decay constant or disintegration constant
$t$ = time taken
Half-life of the given sample,
\[{T_{1/2}} = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.6931}}{\lambda }\]
Where, \[{T_{1/2}}\] = half-life

Complete step by step solution:
The law of radioactive decay says that for every radioactive decay the rate of decay is directly proportional to the number of atoms present at that instance, which gives the law as follows,
\[N = {N_0}{e^{ - \lambda t}}\]
The half-life of a given sample can be said to be the time taken for the sample to decay into one half of its initial value.
\[{T_{1/2}} = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.6931}}{\lambda }\]

To find the half-life of the source, we have to rearrange the above two formulae as,
\[N = {N_0}{e^{ - \dfrac{{\ln 2}}{{{T_{1/2}}}}t}}\]
The exponent and natural log cancels against each other, so,
\[N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{T_{1/2}}}}}}\]
Given, \[{N_0} = 240\], \[N = 30\] and \[t = 60\min \]
To find, \[{T_{1/2}} = ?\]
By using the above formula,
\[30 = 240{\left( {\dfrac{1}{2}} \right)^{\dfrac{{60}}{{{T_{1/2}}}}}} \\ \]
\[\Rightarrow \dfrac{{30}}{{240}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{60}}{{{T_{1/2}}}}}} \\ \]
\[\Rightarrow \dfrac{1}{8} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{60}}{{{T_{1/2}}}}}} \\ \]
\[\Rightarrow {\left( {\dfrac{1}{2}} \right)^3} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{60}}{{{T_{1/2}}}}}} \\ \]
Since the bases are the same, we can equate the powers.
\[3 = \dfrac{{60}}{{{T_{1/2}}}} \\ \]
\[\Rightarrow {T_{1/2}} = \dfrac{{60}}{3} \\ \]
\[\therefore {T_{1/2}} = 20\min \]

Hence, the correct answer is option D.

Note: In the above case, when the exponent and the natural log are inverse functions, they cancel each other and the natural log has a negative sign, hence 2 becomes the denominator of the equation with the remaining terms raised to its power.