Verified Solution • 2025–26 Session moderate

QPages 442-444: Q24: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q24 Solution: The count rate from a radioactive sample falls from per second…

⚠️ Common Exam Traps & Conceptual Pitfalls

When the activity drops from $4.0 imes10^6$ to $1.0 imes10^6$ in 20 h, the ratio is $1/4=(1/2)^2$, so there are two half‑lives, not four. Thus the half‑life is $T=20/2=10$ h. After 100 h you have $100/10=10$ half‑lives, so $A=4.0 imes10^6/2^{10}approx3.9 imes10^3$ s$^{-1}$. Avoid dropping the exponent or miscounting the number of half‑lives.

The count rate from a radioactive sample falls from 4.0×1064.0 \times 10^6 per second to 1.0×1061.0 \times 10^6 per second in 20 hours. What will be the count rate 100 hours after the beginning?

Step-by-Step Verified Solution

**Answer:** Count rate at t = 100 hours = 3.9 _ 10_ per second. **Step-by-Step Explanation:** Given: A_ = 4.0 _ 10_/s, A = 1.0 _ 10_/s at t = 20 h. A/A_ = 1/4 = (1/2)_ _ 2 half-lives in 20 h _ T_/_ = 10 h. At t = 100 h = 10 half-lives: A = A_ (1/2)__ = 4.0 _ 10_ / 1024 = 3.9 _ 10_ per second[citation:9].

All Questions in this Chapter

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