
A radioactive sample is undergoing \[\alpha \]decay. At any time \[{t_1}\], its activity is A and at another time \[{t_2}\], the activity is \[\dfrac{A}{5}\] . What is the average lifetime for the sample?
A. \[\dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]
B. \[\dfrac{{\ln \left( {{t_2} + {t_1}} \right)}}{2}\]
C. \[\dfrac{{\left( {{t_1} - {t_2}} \right)}}{{\ln 5}}\]
D. \[\dfrac{{\ln 5}}{{\left( {{t_2} - {t_1}} \right)}}\]
Answer
134.7k+ views
Hint:Radioactive decay is defined as the emission of energy in the form of ionizing radiation. It involves the spontaneous transformation of one element into another. This can happen only by changing the number of protons in the nucleus.
Formula Used:
To find the activity of the radioactive sample the formula is,
\[A = {A_0}{e^{ - \lambda t}}\]
Where, \[{A_0}\] is initial activity, \[\lambda \] is decay constant and \[t\] is the half-life of a decaying substance.
Complete step by step solution:
Consider a radioactive sample which undergoes \[\alpha \] decay. At any time \[{t_1}\], its activity is A and at another time \[{t_2}\], the activity is \[\dfrac{A}{5}\] . We need to find the average lifetime for the sample. We have studied that the activity of the radio sample is,
\[A = {A_0}{e^{ - \lambda t}}\]
That is the activity decreases exponentially with time.
Here, the activity at time \[{t_1}\]is,
\[{A_1} = {A_0}{e^{ - \lambda {t_1}}}\]……….. (1)
Similarly, the activity at time \[{t_2}\]is,
\[{A_2} = {A_0}{e^{ - \lambda {t_2}}}\]………….. (2)
Here they have given, \[{A_1} = A\]and \[{A_2} = \dfrac{A}{5}\]
Then, the equation (1) and (2), we get,
\[A = {A_0}{e^{ - \lambda {t_1}}}\]……… (3)
\[\Rightarrow \dfrac{A}{5} = {A_0}{e^{ - \lambda {t_2}}}\]……….. (4)
Now, divide the equation (3) by (4), we get,
\[\dfrac{A}{{\dfrac{A}{5}}} = \dfrac{{{A_0}{e^{ - \lambda {t_1}}}}}{{{A_0}{e^{ - \lambda {t_2}}}}}\]
\[\Rightarrow 5 = \dfrac{{{e^{ - \lambda {t_1}}}}}{{{e^{ - \lambda {t_2}}}}}\]
\[\Rightarrow 5 = {e^{\lambda \left( {{t_2} - {t_1}} \right)}}\]
To eliminate the exponential term, we will take the natural logarithm on both sides, that is,
\[\lambda \left( {{t_2} - {t_1}} \right) = \ln 5\]
\[\Rightarrow \dfrac{1}{\lambda } = \dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]
The mean or average lifetime is,
\[\tau = \dfrac{1}{\lambda }\]
\[\therefore \tau = \dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]
Therefore, the average lifetime for the sample is \[\dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\].
Hence, Option A is the correct answer
Note:Remember that whenever there is an exponential term, in order to solve this, we need to take the natural logarithm on both sides, then we can resolve it easily.
Formula Used:
To find the activity of the radioactive sample the formula is,
\[A = {A_0}{e^{ - \lambda t}}\]
Where, \[{A_0}\] is initial activity, \[\lambda \] is decay constant and \[t\] is the half-life of a decaying substance.
Complete step by step solution:
Consider a radioactive sample which undergoes \[\alpha \] decay. At any time \[{t_1}\], its activity is A and at another time \[{t_2}\], the activity is \[\dfrac{A}{5}\] . We need to find the average lifetime for the sample. We have studied that the activity of the radio sample is,
\[A = {A_0}{e^{ - \lambda t}}\]
That is the activity decreases exponentially with time.
Here, the activity at time \[{t_1}\]is,
\[{A_1} = {A_0}{e^{ - \lambda {t_1}}}\]……….. (1)
Similarly, the activity at time \[{t_2}\]is,
\[{A_2} = {A_0}{e^{ - \lambda {t_2}}}\]………….. (2)
Here they have given, \[{A_1} = A\]and \[{A_2} = \dfrac{A}{5}\]
Then, the equation (1) and (2), we get,
\[A = {A_0}{e^{ - \lambda {t_1}}}\]……… (3)
\[\Rightarrow \dfrac{A}{5} = {A_0}{e^{ - \lambda {t_2}}}\]……….. (4)
Now, divide the equation (3) by (4), we get,
\[\dfrac{A}{{\dfrac{A}{5}}} = \dfrac{{{A_0}{e^{ - \lambda {t_1}}}}}{{{A_0}{e^{ - \lambda {t_2}}}}}\]
\[\Rightarrow 5 = \dfrac{{{e^{ - \lambda {t_1}}}}}{{{e^{ - \lambda {t_2}}}}}\]
\[\Rightarrow 5 = {e^{\lambda \left( {{t_2} - {t_1}} \right)}}\]
To eliminate the exponential term, we will take the natural logarithm on both sides, that is,
\[\lambda \left( {{t_2} - {t_1}} \right) = \ln 5\]
\[\Rightarrow \dfrac{1}{\lambda } = \dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]
The mean or average lifetime is,
\[\tau = \dfrac{1}{\lambda }\]
\[\therefore \tau = \dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\]
Therefore, the average lifetime for the sample is \[\dfrac{{\left( {{t_2} - {t_1}} \right)}}{{\ln 5}}\].
Hence, Option A is the correct answer
Note:Remember that whenever there is an exponential term, in order to solve this, we need to take the natural logarithm on both sides, then we can resolve it easily.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

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

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

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

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

JEE Advanced 2024 Syllabus Weightage

JEE Main Chemistry Question Paper with Answer Keys and Solutions
