Atoms and Nuclei JEE 2024: Uranium–Lead Age Calculation

What is the answer to the Atoms and Nuclei JEE 2024 uranium–lead question?

The supplied solution’s answer to this Atoms and Nuclei JEE 2024 question is approximately 142.65, obtained by converting the mass ratio into an atom ratio and then rounding the population ratio. This is a JEE Advanced 2024, Paper 2, Physics numerical-answer question, not an MCQ.

A sample initially contains only U-238. Some uranium decays into Pb-206, leaving a mixture of surviving uranium and lead produced by decay. The lead mass is now seven times the remaining uranium mass. Find the sample’s age using the supplied half-life and logarithm, in the requested format:

t = P \times 10^{8} years, t_{1 / 2} = 4.5 \times 10^{9} years, \ln 2 = 0.693 .

The question bank tags this as medium difficulty, with a 180-second solving target. These are question-bank labels, not official exam classifications.

How do you convert the lead-to-uranium mass ratio into an atom ratio?

Divide each isotope’s mass by its molar mass before comparing populations. Lead-206 atoms are lighter than uranium-238 atoms, so a lead mass seven times the uranium mass means more than seven lead atoms per surviving uranium atom.

Define the present-day counts:

N_{U} = number of remaining U-238 atoms, N_{P b} = number of Pb-206 atoms formed .

In the supplied solution’s model, each decayed uranium atom ultimately produces one lead atom. Using Avogadro’s constant to convert atom counts to moles, the masses are:

m_{P b} = \frac{N_{P b}}{N_{A}} \times 206 g m o l^{- 1}, m_{U} = \frac{N_{U}}{N_{A}} \times 238 g m o l^{- 1} .

Both masses are in grams. Dividing cancels Avogadro’s constant and the common units:

\frac{m_{P b}}{m_{U}} = \frac{206 N_{P b}}{238 N_{U}} = 7 .

Rearranging gives the atom ratio needed for the decay equation:

\frac{N_{P b}}{N_{U}} = \frac{7 \times 238}{206} = \frac{1666}{206} .

How do daughter atoms connect to the radioactive-decay equation?

The initial uranium population equals the surviving uranium count plus the number that decayed into lead. Because the sample starts lead-free, the present lead count tracks decayed uranium atoms under the supplied model.

Define the initial population and write its bookkeeping equation:

N_{0} = initial number of U-238 atoms, N_{0} = N_{U} + N_{P b} .

This is parent-to-final-daughter atom-count bookkeeping, not a claim that uranium and lead masses alone are conserved. The decay chain also releases particles, so equating only those two masses would be wrong.

The exponential decay law gives: $N_{U} = N_{0} e^{- λ t} .$

Rearrange to obtain the initial-to-surviving population ratio:

\frac{N_{0}}{N_{U}} = e^{λ t} .

Substitute the initial population and the converted atom ratio:

e^{λ t} = \frac{N_{U} + N_{P b}}{N_{U}} = 1 + \frac{N_{P b}}{N_{U}} = 1 + \frac{1666}{206} = \frac{1872}{206} .

The added one accounts for surviving uranium atoms, not atoms that produced lead. Taking the reciprocal gives the surviving fraction:

\frac{N_{U}}{N_{0}} = \frac{206}{1872} .

How do you calculate the age and obtain the value of P?

The supplied solution obtains approximately 142.65 by rounding the initial-to-surviving population ratio to nine before taking logarithms. This is an approximation, not an exact equality:

\frac{1872}{206} \approx 9, λ t \approx \ln 9 .

Substitute the half-life relation:

λ = \frac{\ln 2}{t_{1 / 2}}, t \approx \frac{\ln 9}{\ln 2} \times 4.5 \times 10^{9} years .

The supplied numerical working changes logarithm base. The conversion factor cancels between numerator and denominator:

\frac{\ln 9}{\ln 2} = \frac{2 \ln 3}{\ln 2} = \frac{2 \log_{10} 3}{\log_{10} 2} .

Thus, 0.4771 and 0.3010 are common logarithms, not natural logarithms. They are used together as a ratio, rather than mixed with the supplied natural-log value 0.693.

Following the supplied numerical evaluation:

t \approx \frac{2 \times 0.4771}{0.3010} \times 4.5 \times 10^{9} years \approx 14.265 \times 10^{9} years .

Convert the power of ten separately:

14.265 \times 10^{9} = 142.65 \times 10^{8} .

Comparing with the requested age format gives:

P \approx 142.65 .

This is the supplied solution’s approximate result. Retaining the exact population ratio does not produce the identical decimal.

As a scale check, three half-lives leave one eighth of the uranium. Reaching roughly one ninth takes slightly more than three half-lives, consistent with this age. Do not round to an integer unless the answer-entry instructions require it.

Why does using the mass ratio directly give the wrong answer?

Putting seven directly into the population equation gives the wrong result, 135, because it treats unlike isotopes as if their atoms had equal masses. No answer options are supplied for this numerical-answer question, so this is a worked error, not an invented distractor.

The incorrect assumption is:

\frac{N_{P b}}{N_{U}} = 7 (incorrect) .

Following that faulty assumption gives: $e^{λ t} = 1 + 7 = 8,$

t = \frac{\ln 8}{\ln 2} t_{1 / 2} = 3 \times 4.5 \times 10^{9} years .

Converting to the requested format:

t = 13.5 \times 10^{9} years = 135 \times 10^{8} years, P = 135 (wrong) .

The logarithm and power-of-ten conversion are not the problem. The error occurs earlier: omitting the molar-mass factor.

Required correction factor = \frac{238}{206} .

Repair rule: convert each mass to an atom count before using parent–daughter population equations.

What three related questions can you solve using this method?

Practise finding a mass ratio after one half-life, deriving the surviving fraction and finding the age when the isotope masses are equal. These are original exercises, not additional verified PYQs. Each keeps the initially pure uranium sample and the supplied one-parent-to-one-final-daughter model.

What is the lead-to-uranium mass ratio after one half-life?

Half the initial uranium survives and half has produced lead. The atom counts are equal, but their masses are not:

N_{P b} = N_{U} = \frac{N_{0}}{2}, \frac{m_{P b}}{m_{U}} = \frac{206}{238} = \frac{103}{119} .

The mass ratio is less than one because each lead atom is lighter.

What surviving uranium fraction follows from a measured mass ratio?

Define the measured lead-to-uranium mass ratio, then convert it to an atom ratio. Use the initial-population sum to find the surviving fraction:

r = \frac{m_{P b}}{m_{U}}, \frac{N_{P b}}{N_{U}} = \frac{238}{206} r,

\frac{N_{U}}{N_{0}} = \frac{1}{1 + (238 / 206) r} .

How many half-lives have passed when the uranium and lead masses are equal?

Equal masses mean more lead atoms than uranium atoms. Substitute a mass ratio of one into the population relation:

e^{λ t} = 1 + \frac{238}{206}, \frac{t}{t_{1 / 2}} = \frac{\ln (1 + 238 / 206)}{\ln 2} > 1 .

The logarithm’s argument exceeds two, so the age exceeds one half-life.

Rework these without looking at the answers, writing the mass-to-atom conversion before every decay equation.

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Frequently asked questions

What is the answer to the JEE Advanced 2024 uranium–lead age question?

The supplied solution gives P ≈ 142.65, where the sample’s age is t = P × 10^8 years. It obtains this approximate result by rounding the initial-to-surviving uranium population ratio, 1872/206, to nine before taking logarithms. Retaining the exact ratio gives a different decimal.

How do I convert the lead-to-uranium mass ratio into an atom ratio?

Divide each isotope’s mass by its molar mass before comparing atom counts. For Pb-206 and U-238, N_Pb/N_U = (m_Pb/m_U) × (238/206). A lead-to-uranium mass ratio of seven therefore gives an atom ratio of 1666/206, not seven.

Why do we add one to the lead-to-uranium atom ratio?

For an initially pure uranium sample under the one-parent-to-one-final-daughter model, N_0 = N_U + N_Pb. Dividing by the surviving uranium count gives N_0/N_U = 1 + N_Pb/N_U. The added one accounts for surviving uranium atoms.

Why is P = 135 wrong for the uranium–lead question?

P = 135 comes from incorrectly treating the mass ratio of seven as an atom ratio of seven. That assumption gives an initial-to-surviving population ratio of eight and an age of exactly three half-lives. The missing correction factor is 238/206 because uranium and lead atoms have different masses.