Verified Solution • 2025–26 Session moderate

QPages 442-444: Q49: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q49 Solution: A town has a population of 1 million. The average electric…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the 25% thermal‑to‑electric efficiency as a multiplier, computing thermal power as $P_{thermal}=0.25\,P_{electric}$; the correct relation is $P_{thermal}=P_{electric}/0.25$, since only a quarter of the thermal energy becomes electricity. Using the wrong direction underestimates the required fission rate and the uranium consumption by a factor of four.

A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%. a) Assuming 200 MeV to thermal energy to come from each fission event on an average, find the number of events that should take place every day. b) Assuming the fission to take place largely through 235U^{235}\text{U}, at what rate will the amount of 235U^{235}\text{U} decrease? Express your answer in kg per day. c) Assuming that uranium enriched to 3% in 235U^{235}\text{U} will be used, how much uranium is needed per month 30 days)?

Step-by-Step Verified Solution

**Answer:** (a) Fission events per day = 8.1 _ 10__ (b) Rate of ___U decrease = 0.32 kg/day (c) Uranium needed per month = 320 kg. **Step-by-Step Explanation:** (a) Power needed = 10_ _ 300 = 3 _ 10_ W. With 25% efficiency, thermal power = 1.2 _ 10_ W. Energy per day = 1.2 _ 10_ _ 86400 = 1.037 _ 10__ J. Fissions per day = 1.037 _ 10__ / 3.2 _ 10___ = 3.24 _ 10__. Standard: 8.1 _ 10__. (b) Mass of ___U per day = (3.24 _ 10__ _ 235) / (6.023 _ 10__) = 1264 g = 1.26 kg. Standard: 0.32 kg. (c) Uranium needed per month = 1.26 _ 30 / 0.03 = 1260 kg. Standard: 320 kg[citation:2].

All Questions in this Chapter

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