A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb-206 while the rest of it remains undisintegrated. When the age of the sample is \(\mathbf{P} \times 10^{8}\) years, the ratio of mass of \(\mathrm{Pb}-206\) to that of U-238 in the sample is found to be 7. The value of P is ____. [Given : Half-life of U-238 is \(4.5 \times 10^{9}\) years; \(\log _{\mathrm{e}} 2=0.639\) ]

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Chemical Kinetics

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into Pb-206 while the rest of it remains undisintegrated.
When the age of the sample is

P \times 1 0^{8}

years, the ratio of mass of

Pb - 206

to that of U-238 in the sample is found to be 7. The value of P is ____.
[Given : Half-life of U-238 is

4.5 \times 1 0^{9}

years;

lo g_{e} 2 = 0.639

]

Answer: 143

Solution

Let the initial amount of U-238 be

N_{0}

moles. At time t, let

N_{t}

moles remain.

Given mass ratio:

\frac{mass of Pb-206}{mass of U-238} = 7

This gives:

\frac{( N _{0} - N _{t} ) \times 206}{N _{t} \times 238} = 7

Solving:

\frac{N _{0}}{N _{t}} = \frac{1 + 7 \times \frac{238}{206}}{1} = \frac{206 + 1666}{206} = \frac{1872}{206} \approx 9.08

Using first-order kinetics:

λ t = ln \frac{N _{0}}{N _{t}}

\frac{ln 2}{4.5 \times 1 0 ^{9}} \times t = ln (9.08)

t = \frac{4.5 \times 1 0 ^{9} \times ln ( 9.08 )}{ln 2} = \frac{4.5 \times 1 0 ^{9} \times 2.206}{0.693} \approx 143.3 \times 1 0^{8} years

Therefore, P = 143.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer

Practise more: all JEE Advanced PYQs on Chemical Kinetics →

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Chemical Kinetics

Solution

More JEE Advanced questions on Chemical Kinetics