
A sample contains 16 gm of radioactive material, the half-life of which is two days. After 32 days, the amount of radioactive material left in the sample is:
A. Less than 1 mg
B. $\dfrac{1}{2}~gm$
C. $$\dfrac{1}{4}~gm$$
D. 1 gm
Answer
161.4k+ views
Hint: The basic concept of this question is the exponential decay law. Using the half-life equation, calculate the value of the decay constant. Further substituting the value of the decay constant as well as given values in the formula used, determine N which will ultimately give the required solution.
Formula used:
$N=N_{0}e^{-\Lambda~t}$
$T_{1/2}=\dfrac{\log_{e}{2}}{\Lambda}$
Complete step by step solution:
As stated in the question, the half-life of radioactive material is 2 days.
The exponential radioactive law can be expressed mathematically as follows:
$N=N_{0}e^{-\Lambda~t}$..........(1)
In addition, one of the half-life equations is,
$T_{1/2}=\dfrac{\log_{e}{2}}{\Lambda}$.........(2)
It is given that $T_{1/2}=2~days$..........(3)
By combining equations (2) and (3) we get
$T_{1/2}=\dfrac{\log_{e}{2}}{\Lambda}=2~days$
$\Rightarrow\Lambda=\dfrac{\log_{e}{2}}{2}$
Also, it is given that t=32 days
After substituting the value of t and $\Lambda$ in equation (1) we get
$N=N_{0}~e^{-\dfrac{\log_{e}{2}}{2}\times32}$
$\Rightarrow~N=16~e^{-16{\log_{e}{2}}}$.......(4)
Let,
$e^{-16{\log_{e}{2}}}=y$
Now, taking $\log_{e}$ on both sides,
$\log_{e}e^{-16{\log_{e}{2}}}=\log_{e}{y}$
Now, the power of e will appear in the front.
$-16\log_{e}{2}\log_{e}{e}=\log_{e}{y}$
We know $\log_{e}{e}=1$
Now,
$-16\log_{e}{2}=\log_{e}{y}$
Let's take -3 in the power of 2.
$\log_{e}{2^{-16}}=\log_{e}{y}$
After canceling the log on both sides, we get
$2^{-16}=y$
This results in,
$\lgroup\dfrac{1}{2}\rgroup^{16}=y$
Let’s substitute the value of y in equation (4)
$N=16~\lgroup\dfrac{1}{2}\rgroup^{16}$
$\Rightarrow N=2^{4}\times2^{-16}=2^{-12}$
$\Rightarrow N=\lgroup\dfrac{1}{2}\rgroup^{16}$........(5)
Equation (5) is less than 1mg.
Hence, the correct option is A.
Note: Another way to answer this question is to use the exponential decay law equation, which is irrespective of the decay constant. The equation is as follows: $N=N_{0}\lgroup\dfrac{1}{2}\rgroup^{\dfrac{t}{T}}$. We may simply obtain N by applying this equation.
Formula used:
$N=N_{0}e^{-\Lambda~t}$
$T_{1/2}=\dfrac{\log_{e}{2}}{\Lambda}$
Complete step by step solution:
As stated in the question, the half-life of radioactive material is 2 days.
The exponential radioactive law can be expressed mathematically as follows:
$N=N_{0}e^{-\Lambda~t}$..........(1)
In addition, one of the half-life equations is,
$T_{1/2}=\dfrac{\log_{e}{2}}{\Lambda}$.........(2)
It is given that $T_{1/2}=2~days$..........(3)
By combining equations (2) and (3) we get
$T_{1/2}=\dfrac{\log_{e}{2}}{\Lambda}=2~days$
$\Rightarrow\Lambda=\dfrac{\log_{e}{2}}{2}$
Also, it is given that t=32 days
After substituting the value of t and $\Lambda$ in equation (1) we get
$N=N_{0}~e^{-\dfrac{\log_{e}{2}}{2}\times32}$
$\Rightarrow~N=16~e^{-16{\log_{e}{2}}}$.......(4)
Let,
$e^{-16{\log_{e}{2}}}=y$
Now, taking $\log_{e}$ on both sides,
$\log_{e}e^{-16{\log_{e}{2}}}=\log_{e}{y}$
Now, the power of e will appear in the front.
$-16\log_{e}{2}\log_{e}{e}=\log_{e}{y}$
We know $\log_{e}{e}=1$
Now,
$-16\log_{e}{2}=\log_{e}{y}$
Let's take -3 in the power of 2.
$\log_{e}{2^{-16}}=\log_{e}{y}$
After canceling the log on both sides, we get
$2^{-16}=y$
This results in,
$\lgroup\dfrac{1}{2}\rgroup^{16}=y$
Let’s substitute the value of y in equation (4)
$N=16~\lgroup\dfrac{1}{2}\rgroup^{16}$
$\Rightarrow N=2^{4}\times2^{-16}=2^{-12}$
$\Rightarrow N=\lgroup\dfrac{1}{2}\rgroup^{16}$........(5)
Equation (5) is less than 1mg.
Hence, the correct option is A.
Note: Another way to answer this question is to use the exponential decay law equation, which is irrespective of the decay constant. The equation is as follows: $N=N_{0}\lgroup\dfrac{1}{2}\rgroup^{\dfrac{t}{T}}$. We may simply obtain N by applying this 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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

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

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

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

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
