Verified Solution • 2025–26 Session moderate

QPages 442-444: Q46: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q46 Solution: Radioactive isotopes are produced in a nuclear physics…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that the ratio $i/N$ is constant simply because both quantities start at zero, failing to recognize that the ratio of two functions of the form $1 - e^{-kt}$ is constant only if their decay constants $k$ are identical. You must explicitly equate the circuit time constant $R/L$ with the radioactive decay constant $\lambda$ to solve for the half-life.

Radioactive isotopes are produced in a nuclear physics experiment at a constant rate dN/dt=RdN/dt = R. An inductor of inductance 100 mH, a resistor of resistance 100 Ω100 \ \Omega and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that i/Ni/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope.

Step-by-Step Verified Solution

**Answer:** Half-life = 0.693 _ 10__ s. **Step-by-Step Explanation:** The current in the LR circuit: i = i_(1 - e^(-Rt/L)). The number of active nuclei: N = (R/_)(1 - e^(-_t)). For i/N to be constant: the time constants must be equal. Rt/L = _t _ _ = R/L = 100/0.1 = 1000 s__. T_/_ = 0.693/1000 = 6.93 _ 10__ s[citation:2].

All Questions in this Chapter

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