Reactor Heat Lab — how fast does a liquid warm as a radioactive nuclide builds up? — JEE Advanced 2026 Paper 2 Q.2

01The question

Q.2 A nuclear reactor starts producing a radioactive nuclide X from t = 0, at a constant rate of α per second. Each decay of X produces energy E₀, which is utilized to heat a liquid of mass m and specific heat s. Assuming no heat loss from the liquid and taking λ as the decay constant of X, the rate of increase in the temperature of the liquid is:

(A)αE₀m s(1 − e^−λt)
(B)αE₀m s(e^λt − 1)
(C)λE₀m s(1 − e^−λt)
(D)E₀m s(α − λe^−λt)

02The experiment

Target:rate of temperature rise dT/dt

Ready.

Production rate α2.0 × 10¹² s⁻¹

Half-life of X2.0 s

Liquid mass m0.20 kg

Speed

03Detailed solution

Answer: dT/dt = (αE₀/ms)(1 − e^−λt)

Count the nuclei. Let N(t) be the number of nuclei of X present. They are made at the rate α and decay at the rate λN, so dN/dt = α − λN, with N(0) = 0.
Solve it. Separating, dN/(α − λN) = dt, so −(1/λ) ln[(α − λN)/α] = t, which gives N(t) = (α/λ)(1 − e^−λt).
Decays per second. The activity is λN = α(1 − e^−λt): zero at the start, rising towards α, when the nuclei decay as fast as they are made.
Heat. Every decay gives E₀ to the liquid and none is lost, so the heating power is P = E₀λN and ms dT/dt = P.
Rate of temperature rise. dT/dt = E₀λN/(ms) = (αE₀/ms)(1 − e^−λt): option (A).
Check the limits. At t = 0 nothing has been made yet, so nothing can decay: dT/dt = 0. After many half-lives the nuclei decay as fast as they are made, α per second, so dT/dt → αE₀/(ms). Only (A) does both.
The other options. (B) grows without limit, as if each nucleus made more nuclei; a steady production balanced by decay can never do that. (C) has λ where the production rate α belongs, so it has the right shape but the wrong size. (D) gives E₀(α − λ)/(ms) at t = 0, heating before a single nucleus exists.

Understand the concept

A radioactive nuclide that is produced at a steady rate does not pile up for ever. The more nuclei there are, the more decay every second, so the number present climbs and then levels off at α/λ, where decay exactly balances production (the saturation level). The heat released follows the decays, not the production, so the liquid starts warming slowly and approaches a steady rate of rise of αE₀/(ms). It reaches about 97 % of that after five half-lives. The same balance explains why a reactor keeps producing "decay heat" after it is switched off: the nuclei already made go on decaying.

Key takeaways

Production at rate α with decay constant λ: dN/dt = α − λN and N = (α/λ)(1 − e^−λt).
Heat follows the activity λN, not the production rate: P = E₀λN.
Steady state (equilibrium): activity = production rate, N = α/λ.
Check every candidate formula at t = 0 and as t → ∞ before trusting it.

Verification: an independent derivation; a symbolic solution of the rate equation; a Monte Carlo of individual nuclei made and decaying at random; a convolution of the production with the decay law; a check that the total heat equals E₀ times the number of decays; every option audited at t = 0, at large t and in size; and the page's own engine, run by tests/verify_p2phyq02.py. The MathonGo solution was read first only for the approach. The official final answer key (JEE Advanced 2026, Paper 2), read last, gives (A), in agreement. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded.

How to use this simulation

Press ▶ START EXPERIMENT. The reactor starts making nuclei of X, which collect in the tank and decay at random.
Each decay sends a packet of energy into the liquid. Watch the counters: made = still in the tank + decayed.
The graph plots dT/dt measured from the thermometer. It starts at zero and levels off when decay balances production.
At the end the four options are drawn over the measured points, and the one that fits is found. Then try What if… switch the reactor off, or change α, the half-life or m.
Replay runs it again. Reset clears it. Classroom fills the screen for a projector.