
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
162k+ views
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.
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Wheatstone Bridge for JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
