Verified Solution • 2025–26 Session moderate

QPages 442-444: Q35: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q35 Solution: The count rate of nuclear radiation coming from a radiation…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the decay equation $R = R_0 e^{-\lambda t}$ with $R_0 = R e^{-\lambda t}$, leading them to plot $\ln(R/R_0)$ instead of $\ln(R_0/R)$. Since $\ln(R/R_0) = -\lambda t$, using the inverted ratio results in a negative slope, causing the decay constant $\lambda$ to be calculated as a negative value, which is physically impossible for a decay process.

The count rate of nuclear radiation coming from a radiation coming from a radioactive sample containing 128I^{128}\text{I} varies with time as follows. Time t minute): 0, 25, 50, 75, 100; Count rate R 109 s−110^9 \text{ s}^{-1}): 30, 16, 8.0, 3.8, 2.0. a) Plot ln⁡R0/R)\ln R_0/R) against t. b) From the slope of the best straight line through the points, find the decay constant λ\lambda. c) Calculate the half-life t1/2t_{1/2}.c) Calculate the half-life t1/2t_{1/2}.

Step-by-Step Verified Solution

**Answer:** (a) Plot ln(R_/R) vs t (b) _ = 0.027 min__ (c) T_/_ = 25.7 min. **Step-by-Step Explanation:** (b) From the data: at t=0, R=30; t=25, R=16; t=50, R=8; t=75, R=3.8; t=100, R=2. ln(R_/R) at t=25: ln(30/16)=0.628. At t=50: ln(30/8)=1.322. At t=75: ln(30/3.8)=2.066. At t=100: ln(30/2)=2.708. Slope = (2.708-0.628)/(100-25) = 2.08/75 = 0.0277 min__. _ = 0.0277 min__. (c) T_/_ = 0.693/0.0277 = 25.0 min[citation:2].

All Questions in this Chapter

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