Question

In an old rock, the mass ratio of ${}_{92}^{238}U\,to\,{}_{82}^{206}Pb$ is found to be $595:103$ . The age of the rock is: (Mean life of ${}_{92}^{238}U\,$ is ${T_0}$ ).
A) ${T_0}\,\ln \,1.2$
B) ${T_0}\,\ln \,\dfrac{{698}}{{595}}$
C) ${T_0}\,\,\dfrac{{\ln \,1.2}}{{\ln \,2}}$
D) ${T_0}\,\,\dfrac{{\ln \,\dfrac{{698}}{{595}}}}{{\ln \,2}}$

Answer 1

Hint: We will use the relation of time, constant and concentration in which if we put all the values we get the age of the rock. Here the mass ratio is given because with the help of which we can find out the concentration. Convert concentration into moles by dividing the mass with the molar mass of each element.

Complete solution:
In the question it is given that the ratio of ${}_{92}^{238}U\,to\,{}_{82}^{206}Pb$ is $595:103$ . Mean life of ${}_{92}^{238}U\,$ is given ${T_0}$ instead of any numerical value, we have to find out the age of rock, how old it is. Now we know that mean life is shown by tou $\tau $ and related to $\lambda $ as:
$\tau = \,\dfrac{1}{\lambda }$
So, according to the question as it is given that mean life is having value ${T_0}$ so, we can write the formula as per, ${T_0} = \,\dfrac{1}{\lambda }$ . We know that the initial amount in radiochemistry is shown by the symbol ${N_0}$ and the final amount is shown by $N$ . As we get to know by the question that lead is formed from uranium, we can say that the initial amount ${N_0}$ is equal to the sum of uranium as well as lead while amount at any temperature T is given as only uranium. We can write it in mathematical form as ${N_0} = \,U + Pb$ and $N = \,U$ .
So, as the mass ratio is given in question that ${}_{92}^{238}U\,to\,{}_{82}^{206}Pb$ is $595:103$ , it means that if mass of uranium is given as $595$ the mass of lead will be $103$ now let’s put it in the formula and find out the initial moles. ${N_0} = \,\dfrac{{595}}{{238}} + \dfrac{{103}}{{206}}$ and $N = \,\dfrac{{595}}{{238}}$
We have the formula between $\lambda $ and moles as: $t = {T_0} \times 2.303 \times \,\log \,\dfrac{{\left( {\dfrac{{595}}{{238}} + \dfrac{{103}}{{206}}} \right)}}{{\left( {\dfrac{{595}}{{238}}} \right)}}$
Or $t = \dfrac{1}{\lambda } \times 2.303 \times \,\log \,\dfrac{{{N_0}}}{N}$
Now putting values in the equation as- $t = {T_0} \times 2.303 \times \,\log \,\dfrac{{\left( {\dfrac{{595}}{{238}} + \dfrac{{103}}{{206}}} \right)}}{{\left( {\dfrac{{595}}{{238}}} \right)}}$
In the equation we have putted the value of ${N_0}$ and $N$ , put ${T_0} = \,\dfrac{1}{\lambda }$
If we solve it further by diving and changing log into ln terms we get, $t = {T_0}\ln \,\,\dfrac{{698}}{{595}}$ . It means we get the age of the rock. Thus option B. is correct.

Note: we should always convert the mass in moles before putting in the formula. Solve it by dividing and multiplying, don’t forget to change the log into ln by multiplying the log with $2.303$ , by it the expression gets converted into ln with base e. This conversion is done because in all the options we have given answers in terms.