JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Modern Physics

Note — this is a wording issue, not a physics one. The phrase *'the rate of converting these ${}^4$He to ${}^{12}$C'* can be read two ways: as the number of ${}^4$He atoms consumed per second (which is $3\times$ the reaction rate), or as the rate of the conversion *process* itself, i.e. the rate at which ${}^{12}$C is formed. The underlying physics is identical either way; only the counting differs by a factor of 3. NTA's official key takes the second reading, so we follow it: $n=5$. \n\nEnergy released per reaction: $Q=(3m_{{}^{4}He}-m_{{}^{12}C})c^2=(3\times 4.0026-12)(931.5\text{ MeV})=7.266$ MeV. Reaction rate $=\frac{\text{power}}{Q}=\frac{5.808\times 10^{30}}{7.266\times 10^6\times 1.6\times 10^{-19}}\approx 5\times 10^{42}{\text{s}}^{-1}$. Taking this as the rate of conversion (per the NTA reading), $n=5$.