Question

The activity of a radioactive sample falls from 700 to 500 in 30 minutes. Its half life is close to:
A. 52min
B. 62min
C. 72min
D. 66min

Answer 1

In order to solve this question, we must know the equation of radioactivity. If you know the equation of radioactivity, you can see that to proceed further you must know the basic rules of logarithmic addition and subtraction. We know that every radioactive element undergoes decay. So half-life of a radioactive substance is the time taken by the substance to reach its half value. To know that the substance attains its half-life we should be given either the initial mass of substance or the mass remaining after half life and time required to attain half-life.

Formula used: ${{A}_{t}}={{A}_{0}}{{e}^{-\lambda t}}$
Where,
${{A}_{t}}$ = amount of the material at time t,
${{A}_{0}}$ = amount of material at time t=0,
$\lambda$ = decay constant.

Complete step by step answer:
Half life is calculated for a radioactive material. Since, it is very hard to tell that when a radioactive material will undergo complete decay, we find the half life of such materials.
From the equation of radioactivity,
${{A}_{t}}={{A}_{0}}{{e}^{-\lambda t}}$
After taking log on both sides,
We get,
$\ln \left[ \dfrac{{{A}_{t}}}{{{A}_{0}}} \right]=-\lambda t$
By applying logarithmic rule,
We get,
$\ln \left[ \dfrac{{{A}_{0}}}{{{A}_{t}}} \right]=\lambda t$ ……… (1)
Now, we have ${{A}_{0}}$ = amount of material at time t=0
So, at time t = ${{t}_{1/2}}$, the amount will be half of the initial amount of the material.
Therefore,
${{A}_{t}}=\dfrac{{{A}_{0}}}{2}$ ……………. (2)
After substituting (1) in (2) and t = ${{t}_{1/2}}$
We get,
$\ln 2=\lambda {{t}_{1/2}}$ …………….. (3)
Now, we substitute the given conditions in (1)
We get,
$\ln \left[ \dfrac{700}{500} \right]=\lambda \times 30$ ………….. (4)
Now, from (3) and (4)
We get,
$\dfrac{\ln 2}{\ln \left[ 7/5 \right]}=\dfrac{{{t}_{1/2}}}{30}$
$\Rightarrow {{t}_{1/2}}=\dfrac{\ln 2\times 30}{\ln \left[ 7/5 \right]}$
$\Rightarrow {{t}_{1/2}}=\dfrac{0.693\times 30}{0.336}$
Therefore,
${{t}_{1/2}}=61.28$min.

So, the correct answer is “Option B”.

Note: For this question we were not given this answer in our option but we know that our process and our calculation is very accurate. So, we must choose the answer closest to our answer. Also, to solve this question we must know how to calculate the log of the numbers. For such examples, related to radioactivity and decay knowledge about logs is very necessary.