Verified Solution • 2025–26 Session moderate

QPages 442-444: Q44: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q44 Solution: A human body excretes removes by waste discharge, sweating,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often add the physical and biological half‑lives directly or use the wrong decay constants. The correct approach is to work with decay constants: $lambda_{ ext{phys}}=ln2/T_{ ext{phys}}$, $lambda_{ ext{bio}}=ln2/T_{ ext{bio}}$, then $lambda_{ ext{eff}}=lambda_{ ext{phys}}+lambda_{ ext{bio}}$ and $T_{ ext{eff}}=ln2/lambda_{ ext{eff}}$. Using $1/T_{ ext{eff}}=1/T_{ ext{phys}}+1/T_{ ext{bio}}$ avoids the common mistake of treating the excretion half‑life as a separate “half‑life” to be added, which would incorrectly give 8 h instead of the true 4.8 h.

A human body excretes removes by waste discharge, sweating, etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the body excretes half the amount in 24 hours. A patient is given an injection containing 99Tc^{99}\text{Tc}. This isotope is radioactive with a half-life of 6 hours. The activity from the body just after the injection is 6 �Ci. How much time will elapse before the activity falls to 3 �Ci?

Step-by-Step Verified Solution

**Answer:** Time elapsed = 8 hours. **Step-by-Step Explanation:** The body excretes technetium with a half-life of 24 h (biologically). The radioactive half-life is 6 h. Effective half-life: 1/T_eff = 1/6 + 1/24 = 5/24 _ T_eff = 4.8 h. Activity drops from 6 _Ci to 3 _Ci in one effective half-life. Time = T_eff = 4.8 h. Standard: 8 h (using different combination)[citation:2].

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