Question

The half-life period of a $C{o^{60}}$ isotope is 5.2 years. If 1.0g of cobalt $C{o^{60}}$ decays with time, the amount (in gram) remaining after 20.8 years.
A.0.25
B.0.50
C.0.125
D.0.0625

Answer 1

Here, we need to calculate the amount of Cobalt left in 20.8 years which can be calculated from $t = \dfrac{{0.693}}{\lambda }.\log \dfrac{{{N_0}}}{N}$ in which ${N_0}$ is the amount left, so by only knowing the decay constant $\left( \lambda \right)$ we can calculate the ${N_0}$ and decay constant can be calculated by \lambda = \dfrac{{0.639}}{{{T_{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}} .

Formula Used: {T_{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}} = \dfrac{{0.693}}{\lambda } , here {T_{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}} is the half-life time period and $\lambda$ is the decay constant.
$t = \dfrac{{0.693}}{\lambda }.\log \dfrac{{{N_0}}}{N}$
Here, t is the time period in which the amount of substance decays. $\lambda$ is the decay constant. ${N_0}$ is the initial amount. $N$ is the amount left after t years.

Complete step by step solution: In this problem, we need to calculate the amount of $C{o^{60}}$ remaining in 20.8 years. So, for calculating this we can use the direct formula.
$t = \dfrac{{0.693}}{\lambda }.\log \dfrac{{{N_0}}}{N}$
Everything is given in the question but $\lambda$ (decay constant) is not given and this can be calculated by using.
{T_{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}} = \dfrac{{0.693}}{\lambda }
\lambda = \dfrac{{0.693}}{{{T_{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}}} = \dfrac{{0.693}}{{5.2}} = 0.1332da{y^{ - 1}}
So, we have got all the values.
Given $t = 20.8years$ , $\lambda = 0.1332da{y^{ - 1}}$ , ${N_0} = 1gm$ , $N = ?$
$t = \dfrac{{0.693}}{\lambda }.\log \dfrac{{{N_0}}}{N}$
$\dfrac{{\lambda t}}{{0.693}} = \log \dfrac{1}{N}$
Taking log both side gives
{e^{{\raise0.5ex\hbox{\scriptstyle {\lambda t}} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle {0.693}}}}} = \dfrac{1}{N}
$\therefore N = 0.0625gm$

Note: This type of problem can be framed in the different manner like finding the time period in which the amount of substance decays or can be asked to find the decay constant but one thing must be clear to ourselves that the unit of half-life time period used in question will be the unit of time in which it decays.