Question

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 1

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.