A reactor forms a radioactive nuclide Y from X at a constant rate 10¹⁵ nuclei s⁻¹, releasing 100 MeV for each X→Y decay. At steady state each Y→Z decay releases another 50 MeV. The energy-to-electricity conversion efficiency is 25%. Find electrical power in kW.
Physics · Mixed Chapter
JEE Mains

Question

A reactor forms a radioactive nuclide Y from X at a constant rate 10¹⁵ nuclei s⁻¹, releasing 100 MeV for each X→Y decay. At steady state each Y→Z decay releases another 50 MeV. The energy-to-electricity conversion efficiency is 25%. Find electrical power in kW.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Steady-state daughter-decay rate: In steady state the production and decay rates of Y are both 10¹⁵ s⁻¹. Step 2 — Total released energy per completed decay chain: E = 100+50 = 150 MeV. Step 3 — Thermal power: P₁ = 10¹⁵×150×(1.602×10⁻¹³) J/s = 2.403×10⁴ W. Step 4 — Electrical output: P₂ = 0.25×2.403×10⁴ W ≈ 6007.5 W ≈ 6.0 kW. Answer: 6.