Verified Solution • 2025–26 Session moderate

QPages 442-444: Q28: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q28 Solution: decays to by -emission. The resulting is in its excited state…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly apply the extremely short gamma half-life ($10^{-8}$ s) to the time calculation, assuming the gamma emission rate decays independently. You must recognize that the gamma emission rate is governed by the production rate of the excited daughter, which is determined by the much slower parent beta decay half-life (270 days).

57Co^{57}\text{Co} decays to 57Fe^{57}\text{Fe} by β+\beta^+-emission. The resulting 57Fe^{57}\text{Fe} is in its excited state and comes to the ground state by emitting γ\gamma-rays. The half-life of β+\beta^+-decay is 270 days and that of the γ\gamma-emissions is 10−8 s10^{-8} \text{ s}. A sample of 57Co^{57}\text{Co} gives 5.0×1095.0 \times 10^9 gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5×1092.5 \times 10^9 per second?

Step-by-Step Verified Solution

**Answer:** Time = 270 days. **Step-by-Step Explanation:** The gamma emission rate is determined by the beta decay of __Co (since the gamma emission is instantaneous after beta decay). Emission rate drops to half when the number of __Co nuclei halves. T_/_ of beta decay = 270 days. So time = 270 days[citation:2].

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