Question

The analysis of a rock shows the relative number of ${{\text{U}}^{238}}$ and $\text{P}{{\text{b}}^{206}}$ atoms (Pb/U =\[0.\text{25}\]). The age of rock will be:
(A) \[\dfrac{2.303}{0.693}\times 4.5\times {{10}^{9}}\log 1.25\]
(B) \[\dfrac{2.303}{0.693}\times 4.5\times {{10}^{9}}\log 0.25\]
(C) \[\dfrac{2.303}{0.693}\times 4.5\times {{10}^{9}}\log 4\]
(D) \[\dfrac{2.303}{4.5\times {{10}^{9}}}\times 0.693\log 4\]

Answer 1

Hint: Half-life: the time required by a substance to decay the half of its total quantity is known as half-life of that element. Half-life helps in calculating the decay constant for an element.
The half-life of Uranium 238 is found to be \[{{t}_{\dfrac{1}{2}}}=4.5\times {{10} ^ {9}}\]
Formula Used: The formula to calculate the half-life of an atom is given by,
\[{{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2} {n}\]
Where, n is the units decayed per year

Complete step by step solution:
Given: (Pb/U) =0.25
$\therefore \dfrac{U + Pb}{U}=1.25=\dfrac{5}{4}$ (Adding 1 to both sides)
We know that the half-life formula can be written as,
\[{{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2} {n}\]
\[n=\dfrac{\ln 2}{{{t}_{\dfrac{1}{2}}}}\]
\[n=\dfrac{\ln 2} {4.5\times {{10} ^ {9}}}\]
The formula to calculate age of rock is given by,
\[n=\dfrac{2.303}{t}\times \log (\dfrac{U + Pb}{U})\] (Converting natural logarithm to base 10 log)
Equating both the values of n
\[\dfrac{\ln 2} {4.5\times {{10}^{9}}}=n=\dfrac{2.303}{t}\times \log (\dfrac{U + Pb}{U})\]
$\therefore $ Simplifying this equation, we will get,
$t=\dfrac{2.303}{0.693}(4.5\times {{10}^{9}})\log \dfrac{5}{4}$
\[=\dfrac{2.303}{0.693}\times 4.5\times {{10} ^ {9}}\log 1.25\]

Hence, choice (A) is correct.

Note: The age of the rock can be determined by the amount of uranium contained in it as we are aware of the half-life of uranium. The ratio of lead to uranium decay helps us to calculate the precise value for the age of the rock. All the values should be taken to be accurate up to third decimal places to obtain correct values.