Verified Solution • 2025–26 Session moderate

QPages 442-444: Q27: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q27 Solution: The selling rate of a radioactive isotope is decided by its…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat radioactive decay as a linear loss per half‑life, e.g., assuming the activity after 30 days is 0.25 of the original because two half‑lives have passed, then adjusting linearly for the extra days. The correct approach is always exponential: $N = N_0,(1/2)^{t/T_{1/2}}$, so compute $(1/2)^{30/14.3}$ before multiplying by the original price. Avoid any step‑wise subtraction of fractions; use the power law directly.

The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one month old 32P^{32}\text{P} t1/2=14.3 dayst_{1/2} = 14.3 \text{ days}) source if it was originally purchased for 800 rupees?

Step-by-Step Verified Solution

**Answer:** Second-hand rate � 200 rupees. **Step-by-Step Explanation:** Given: T_/_ = 14.3 days, t = 30 days. Fraction remaining = (1/2)^(30/14.3) = (1/2)^2.1 = 0.233. Value = 800 _ 0.233 = 186 � 200 rupees[citation:2].

All Questions in this Chapter

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