Verified Solution • 2025–26 Session moderate

QPages 442-444: Q38: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q38 Solution: A radioactive isotope is being produced at a constant rate in…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently replace the decay constant $\lambda$ with the half‑life $t_{1/2}$, writing $\lambda = t_{1/2}$ and obtaining $N=R t_{1/2}$; the correct relation is $\lambda =\ln2/t_{1/2}$, so the equilibrium number is $N=R/\lambda = R t_{1/2}/\ln2$. Remember to convert half‑life to decay constant before equating production and decay rates.

A radioactive isotope is being produced at a constant rate dN/dt=RdN/dt = R in an experiment. The isotope has a half-life t1/2t_{1/2}. Show that after a time t≫t1/2t \gg t_{1/2} the number of active nuclei will become constant. Find the value of this constant.

Step-by-Step Verified Solution

**Answer:** Constant number N = R/_ = R T_/_/0.693. **Step-by-Step Explanation:** When production rate R equals decay rate _N, equilibrium is reached: dN/dt = R - _N = 0 _ N = R/_. Since _ = 0.693/T_/_, N = R T_/_/0.693[citation:1].

All Questions in this Chapter

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