Question

The half-life of Radon is $3.8$ days. If we start with $100\,g$of sample how much will be left after $15.2$ days $?$
(i) $6.295\,g$
(ii) $6.5\,g$
(iii) $6.0\,g$
(iv) $6.75\,g$

Answer 1

Hint:The half-life $\left( {{t_{\dfrac{1}{2}}}} \right)$ of Radon is given. Use the equation $\lambda \,\, = \,\,\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$ to calculate the decay constant $\lambda $. Then use the equation $t\, = \,\dfrac{{2.303}}{\lambda }\log \,\left( {\dfrac{{{N_ \circ }}}{N}} \right)$$($where the initial amount of sample taken $\left( {{N_ \circ }} \right)$ is given and we calculated the decay constant$)$ to calculate the amount of sample left after $15.2$ days.

Complete step by step solution:
We know that radioactive decay is a first order reaction and they follow the equations
$t\, = \,\dfrac{{2.303}}{\lambda }\log \,\left( {\dfrac{{{N_ \circ }}}{N}} \right)........\left( 1 \right)$ where ${N_ \circ }$ is the initial amount of sample taken, ${N_ \circ }$ is the amount of sample left after time $t$ and $\lambda $ is the decay constant.
$\lambda $ for a first order reaction can be calculated as $\lambda \,\, = \,\,\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}........\left( 2 \right)$ where ${t_{\dfrac{1}{2}}}$ is the half-life of the sample taken.
The half-life of Radon $\left( {{t_{\dfrac{1}{2}}}} \right)\,\,\, = \,\,3.8\,\,days$ $$
Therefore using equation $\left( 2 \right)$ we get
$\lambda \,\, = \,\,\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}\,\, = \,\,\dfrac{{0.693}}{{3.8\,days}}\,\, = \,\,0.182\,\,day{s^{ - 1}}$
Therefore the decay constant $\lambda \, = \,0.182\,day{s^{ - 1}}$.
The initial amount of sample taken $\left( {{N_ \circ }} \right)\,\, = \,\,100\,g$
Time elapsed $\left( t \right)\,\, = \,\,15.2\,\,days$
Using equation $\left( 1 \right)$ we get
$15.2\,\, = \,\,\dfrac{{2.303}}{{0.182}}\log \,\left( {\dfrac{{100}}{N}} \right)$
$ \Rightarrow \,\,\dfrac{{15.2 \times 0.182}}{{2.303}}\,\, = \,\,\log \left( {\dfrac{{100}}{N}} \right)$
$ \Rightarrow \,\,1.201\,\, = \,\,\log \left( {\dfrac{{100}}{N}} \right)$
$ \Rightarrow \,1.201\, = \,\log \left( {100} \right) - \log N$
$ \Rightarrow \,1.201\, = \,2 - \log N$
$ \Rightarrow \,1.201 - 2 = - \log N$
$ \Rightarrow \,\log N\, = \,0.799$
Taking antilog on both sides we get
$N\, = \,6.295\,g$$$
Therefore after $15.2\,days\,\,6.295\,g$of samples will be left.

Hence the correct answer is (i) $6.295\,g$.


Additional information:A material containing unstable nuclei is considered to be radioactive. Since the nuclei is unstable it has high energy. So in order to lose its energy it undergoes disintegration or decay and this process is referred to as radioactive decay. Radioactive decay can be of three types:
$ \bullet \,\,\alpha $ decay
$ \bullet \,\,\beta $ decay
$ \bullet \,\,\gamma $ decay.
All of these involve emission of a particle or photon in order to release its energy.

Note:You can also solve the problem using the standard equation of a first order reaction. The equation is $N\, = \,{N_ \circ }{e^{ - \dfrac{\lambda }{t}}}$ where the terms hold the same significance. While using this equation make sure to use $\ln $, but if you want to use $\log $ use a proper conversion factor in order to convert $\ln $to $\log $.