Question

In an ore containing uranium, the ratio of \[{U^{238}}\] to \[P{b^{206}}\] is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of \[{U^{238}}\]. Take the half-life of \[{U^{238}}\] to be \[4.5 \times {10^9}yr\]
A. \[1.6 \times {10^3}yr\]
B. \[1.5 \times {10^4}yr\]
C. \[1.867 \times {10^9}yr\]
D. \[2 \times {10^5}yr\]

Answer 1

Hint:Before we proceed with the problem let’s see what is given. They have given the half-life of \[{U^{238}}\]and the ratio of \[{U^{238}}\] to \[P{b^{206}}\] .We need to find the age of the ore. For that first write the formula to find the number of atoms undecayed after time t then by using the ratio of uranium and lead we will calculate the age of the ore.

Formula Used:
The formula to find the law of radioactive disintegration is,
\[N = {N_0}{e^{ - \lambda t}}\]……… (1)
Where, \[N\] is the number of atoms remaining undecayed after time t and \[{N_0}\] is the initial number of atoms.

Complete step by step solution:
Here, at time \[t = 0\], the number of atoms present in the \[{U^{238}}\] is \[{N_0}\] and the number of atoms in \[P{b^{206}}\] is zero. After \[t = t\], the number atoms left in \[{U^{238}}\] is \[N\] and the number of atoms left in the \[P{b^{206}}\] is \[{N_0}\]. They have given that, the ratio of U-238 to Pb-206 is 3. So we can write it as,
\[\dfrac{N}{{{N_0} - N}} = 3\]
\[ \Rightarrow N = \dfrac{3}{4}{N_0}\]…….. (2)

Equating equations (1) and (2) we get,
\[N = {N_0}{e^{ - \lambda t}}\]
\[ \Rightarrow N = \dfrac{3}{4}{N_0}\]
\[\Rightarrow {e^{ - \lambda t}} = \dfrac{3}{4}\]
Taking ln on both sides we get,
\[\ln \left( {{e^{ - \lambda t}}} \right) = \ln \left( {\dfrac{3}{4}} \right)\]
\[\Rightarrow \lambda t = \ln \left( {\dfrac{4}{3}} \right)\]
\[\Rightarrow t = \dfrac{{\ln \left( {\dfrac{4}{3}} \right)}}{\lambda }\]………. (3)

We know that from half- life,
\[\lambda = \left( {\dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}} \right)\]
Put the value of \[\lambda \]in equation (3) we get,
\[t = \dfrac{{\ln \left( {\dfrac{4}{3}} \right)}}{{\ln 2}} \times {T_{\dfrac{1}{2}}}\]
\[\Rightarrow t = \dfrac{{0.288}}{{0.693}} \times 4.5 \times {10^9}\]............(Given: \[{T_{\dfrac{1}{2}}} = 4.5 \times {10^9}yr\])
\[\therefore t = 1.87 \times {10^9}yr\]
Therefore, the age of the ore (Uranium) is \[1.87 \times {10^9}yr\].

Hence, option C is the correct answer.

Note:Half-life of a substance does not depend on the temperature, concentration of a substance, but it depends on the time factor. If the radioactive element is used in a circumstance where half of the atoms have decayed after one-half life, it is reasonable to conclude that they have a well-defined life expectancy, i.e. atoms with a mean life that is substantially greater than their half-life. This means that the mean life equals the half-life divided by 2, which is the natural algorithm.