Question

Two samples of radioactive substances initially contain an equal number of atoms. Their half-life times are respectively for 2 hours and 4 hours. The ratio of their disintegration rates after 12 hours is
A. 1 : 4
B. 1 : 2
C. 3 : 4
D. 2 : 3

Answer 1

This question is based on the concept of the half-life and the disintegration rates of the radioactive samples. We will use the formula of the half-life time to find the quantity of the substances remaining and then we will substitute these values in the disintegration rate equation to find the ratio of the same.

Formula used:
$N={{N}_{0}}{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
$R=\lambda N=\dfrac{0.693N}{{{T}_{{}^{1}/{}_{2}}}}$

Complete step by step answer:
The formula for calculating the half-life of the substance is given as follows
$N={{N}_{0}}{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
Where N is the quantity of the substance remaining, t is the time elapsed, ${{N}_{0}}$is the initial quantity of the substance and ${{t}_{{}^{1}/{}_{2}}}$is the half-life of the substance.
From the data, we have the data, two samples of radioactive substances initially contain an equal number of atoms. Their half-life times are respectively for 2 hours and 4 hours.
So, we need to consider two cases, one for the sample of a radioactive substance having the half-life time to be equal to 2 hours and the other for the sample of a radioactive substance having the half-life time to be equal to 4 hours.
Firstly, consider the radioactive sample having the half-life time to be equal to 2 hours.
The ratio of the radioactive sample after 12 hours is,

& {{N}_{1}}={{N}_{0}}{{\left( \dfrac{1}{2} \right)}^{\dfrac{12}{2}}} \\\ & \Rightarrow {{N}_{1}}=\dfrac{1}{{{2}^{6}}}{{N}_{0}} \\\ \end{aligned}$$ Now, consider the radioactive sample having the half-life time to be equal to 4 hours. The ratio of the radioactive sample after 12 hours is, $$\begin{aligned} & {{N}_{2}}={{N}_{0}}{{\left( \dfrac{1}{2} \right)}^{\dfrac{12}{4}}} \\\ & \Rightarrow {{N}_{2}}=\dfrac{1}{{{2}^{3}}}{{N}_{0}} \\\ \end{aligned}$$ The decay rate is given as follows. $$R=\lambda N=\dfrac{0.693N}{{{T}_{{}^{1}/{}_{2}}}}$$ Where N is the quantity of the radioactive substance remaining and $${{T}_{{}^{1}/{}_{2}}}$$is the half-life of that radioactive substance. $$\Rightarrow R\propto \dfrac{N}{{{T}_{{}^{1}/{}_{2}}}}$$ Again, here also consider two cases. So, we get, The decay rate is the radioactive sample having the half-life time of 2 hours. $${{R}_{1}}\propto \dfrac{{{N}_{1}}}{{{\left( {{T}_{{}^{1}/{}_{2}}} \right)}_{1}}}$$….. (1) The decay rate is the radioactive sample having the half-life time of 4 hours. $${{R}_{2}}\propto \dfrac{{{N}_{2}}}{{{\left( {{T}_{{}^{1}/{}_{2}}} \right)}_{2}}}$$….. (2) Divide the equations (1) and (2), so, we get, $$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{\left( {{T}_{{}^{1}/{}_{2}}} \right)}_{2}}}{{{\left( {{T}_{{}^{1}/{}_{2}}} \right)}_{1}}}\times \dfrac{{{N}_{1}}}{{{N}_{2}}}$$ Substitute the given and obtained values in the above equation, so ,we get, $$\begin{aligned} & \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{4}{2}\times \dfrac{\dfrac{{{N}_{0}}}{{{2}^{6}}}}{\dfrac{{{N}_{0}}}{{{2}^{3}}}} \\\ & \Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=2\times \dfrac{{{2}^{3}}}{{{2}^{6}}} \\\ & \Rightarrow \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{1}{4} \\\ \end{aligned}$$ As, the ratio of their disintegration rates after 12 hours is obtained to be equal to 1 : 4, **Thus, the option (A) is correct.** **Note:** This question is a direct one. We have first used the half-life time of the radioactive sample formula because in order to find the disintegration rates, we need the value of the quantity of the sample remaining. In this case, the ratio of the disintegration rates is asked, even they can ask for the ratio of the radioactive samples remaining by providing the values of the disintegration rates.