Question

The half-life of a sample of a radioactive substance is 1 hour. If \[8\times {10^{10}}\] atoms are present at t = 0, then the number of atoms decayed in the duration t = 2 hour to t = 4 hour will be
A. \[2\times {10^{10}}\]
B. \[1.5\times {10^{10}}\]
C. Zero
D. Infinity

Answer 1

Hint:The law of radioactive decay can calculate approximately how many nuclei in a sample are decayed over a period of time. The number of atoms will decrease exponentially, which means it takes infinite time to completely decay. Half-life can be defined as the time taken for one half of the sample to decay from the initial amount of sample present.

Formula Used:
To find number of atoms decayed during the given time, the law of radioactive decay is,
\[N = {N_0}{e^{ - \lambda t}}\]
Where, \[N\] is the number of atoms present after the given period of time
\[{N_0}\] is the number of atoms present initially
\[\lambda \] is the radioactive decay constant or disintegration constant
\[t\] is the decay time
Half-life of sample is,
\[{T_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.6931}}{\lambda }\]
Where, \[{T_{\dfrac{1}{2}}}\] is the half-life of the given sample

Complete step by step solution:
Given, the number of atoms present initially (at t = 0), \[{N_0} = 8{\rm{x1}}{{\rm{0}}^{10}}\]
The half-life , \[{T_{\dfrac{1}{2}}}\] = 1 hour
To find the number of atoms decayed, N at t =2 to t = 4 hour.
So, \[N = {N_0}{e^{ - \lambda t}}\]
\[\Rightarrow \lambda = \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}} \\ \]
\[\Rightarrow N = {N_0}{e^{ - \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}t}} \\ \]
\[\Rightarrow N = {N_0}{e^{ - \ln 2t}} \\ \]
The exponent and natural log gets cancelled against each other, then
\[N = {N_0}{\left( {\dfrac{1}{2}} \right)^t}\]
At t = 2 hour,
\[N = 8 \times {\rm{1}}{{\rm{0}}^{10}}{\left( {\dfrac{1}{2}} \right)^2} \\ \]
\[\Rightarrow N = 8 \times {\rm{1}}{{\rm{0}}^{10}}\left( {\dfrac{1}{4}} \right) \\ \]
\[\Rightarrow N = 2 \times {\rm{1}}{{\rm{0}}^{10}} \\ \]
At t = 4 hour,
\[N = 8 \times {\rm{1}}{{\rm{0}}^{10}}{\left( {\dfrac{1}{2}} \right)^4} \\ \]
\[\Rightarrow N = 8\times {\rm{1}}{{\rm{0}}^{10}}\left( {\dfrac{1}{{16}}} \right) \\ \]
\[\Rightarrow N = \dfrac{1}{2} \times {\rm{1}}{{\rm{0}}^{10}} \\ \]
So we have the number of atoms present at t = 2 hour as \[N = 2\times {\rm{1}}{{\rm{0}}^{10}}\] and at t = 4 hours as \[N = \dfrac{1}{2} \times {\rm{1}}{{\rm{0}}^{10}}\]. To find the number of atoms present during t = 2 hours to t = 4 hours is,
\[N = \left( {2 - \dfrac{1}{2}} \right)\times {\rm{1}}{{\rm{0}}^{10}}\]
\[\therefore N = 1.5 \times {\rm{1}}{{\rm{0}}^{10}}\]

Hence, the correct answer is option B.

Note: It is asked to find the number of atoms during the duration of t = 2 hours and t = 4 hours, so we have calculated N for both separately and found the difference of both to find N in the duration. Remember when the exponent and natural log were cancelled against each other, the remaining 2 is negative, hence it is written in the denominator of the equation.