
The half-life of a radioactive substance is 40 years. How long will it take to reduce to one fourth of its original amount and what is the value of decay constant?
A. 40 years, 0.9173/year
B. 90 years, 9.017/year
C. 80 years, 0.0173/year
D. None of these
Answer
162.3k+ views
Hint:The half-life of a radioactive substance can be described as the time needed for it to reduce into one-half of the initial amount. Half-life is the common way to characterize the rate of decay of radioactive substances, since they vary from each other.
Formula used :
The half-life of a radioactive substance is,
\[{T_{1/2}} = \dfrac{{0.6931}}{\lambda }\]
Where, \[{T_{1/2}}\] is the half-life and \[\lambda \] is the decay or disintegration constant.
Complete step by step solution:
Given, \[T_{1/2} = 40years\]
To find, \[{T_{1/4}} = ?\]
\[\lambda = ?\]
To find one-fourth of the substance, consider,
\[{N_0}\mathop \to \limits^{{T_{1/2}}} \dfrac{{{N_0}}}{2}\mathop \to \limits^{2{T_{1/2}}} \dfrac{{{N_0}}}{4} \\ \]
The half-life of a radioactive substance is,
\[{T_{1/2}} = \dfrac{{0.6931}}{\lambda } \\ \]
So, for the radioactive sample to get one-fourth of its initial amount, it needs two half-lives.
\[{T_{1/4}} = 2{T_{1/2}} \\ \]
\[\Rightarrow {T_{1/4}} = 2 \times {\rm{40}} \\ \]
\[\Rightarrow {T_{1/4}} = 80\,years \\ \]
To find the decay constant,
\[{T_{1/2}} = \dfrac{{0.6931}}{\lambda } \\ \]
\[\Rightarrow \lambda = \dfrac{{0.6931}}{{{T_{1/2}}}} \\ \]
\[\Rightarrow \lambda = \dfrac{{0.6931}}{{40}} \\ \]
\[\therefore \lambda = 0.0173/year\]
The time taken for the radioactive substance to get one-fourth of its initial amount is 80 years and the decay constant is 0.0173/year.
Hence, the correct option is C.
Note: Here, we need to find ,one-fourth of the initial amount of the radioactive substance, hence we calculated it as two half-lives, because during the first half-life the initial amount will be reduced to one-half \[\left( {1/2} \right)\] and when second half-life occurs, the amount of radioactive substance will be reduced by another \[\left( {1/2} \right)\], i,e., \[\left( {\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}} \right)\] which gives us the result.
Formula used :
The half-life of a radioactive substance is,
\[{T_{1/2}} = \dfrac{{0.6931}}{\lambda }\]
Where, \[{T_{1/2}}\] is the half-life and \[\lambda \] is the decay or disintegration constant.
Complete step by step solution:
Given, \[T_{1/2} = 40years\]
To find, \[{T_{1/4}} = ?\]
\[\lambda = ?\]
To find one-fourth of the substance, consider,
\[{N_0}\mathop \to \limits^{{T_{1/2}}} \dfrac{{{N_0}}}{2}\mathop \to \limits^{2{T_{1/2}}} \dfrac{{{N_0}}}{4} \\ \]
The half-life of a radioactive substance is,
\[{T_{1/2}} = \dfrac{{0.6931}}{\lambda } \\ \]
So, for the radioactive sample to get one-fourth of its initial amount, it needs two half-lives.
\[{T_{1/4}} = 2{T_{1/2}} \\ \]
\[\Rightarrow {T_{1/4}} = 2 \times {\rm{40}} \\ \]
\[\Rightarrow {T_{1/4}} = 80\,years \\ \]
To find the decay constant,
\[{T_{1/2}} = \dfrac{{0.6931}}{\lambda } \\ \]
\[\Rightarrow \lambda = \dfrac{{0.6931}}{{{T_{1/2}}}} \\ \]
\[\Rightarrow \lambda = \dfrac{{0.6931}}{{40}} \\ \]
\[\therefore \lambda = 0.0173/year\]
The time taken for the radioactive substance to get one-fourth of its initial amount is 80 years and the decay constant is 0.0173/year.
Hence, the correct option is C.
Note: Here, we need to find ,one-fourth of the initial amount of the radioactive substance, hence we calculated it as two half-lives, because during the first half-life the initial amount will be reduced to one-half \[\left( {1/2} \right)\] and when second half-life occurs, the amount of radioactive substance will be reduced by another \[\left( {1/2} \right)\], i,e., \[\left( {\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}} \right)\] which gives us the result.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

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

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

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

In which of the following forms the energy is stored class 12 physics JEE_Main
