Verified Solution • 2025–26 Session moderate

QPages 442-444: Q53: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q53 Solution: Calculate the energy that can be obtained from 1 kg of water…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently read a value like $1.5\times10^{-2}\%$ as the fraction $1.5\times10^{-2}$ (0.015) instead of converting the percent correctly to $1.5\times10^{-4}$; always divide the scientific‑notation number by 100 before using it as a fraction of the total.

Calculate the energy that can be obtained from 1 kg of water through the fusion reaction 2H+2H→3H+p^2\text{H} + ^2\text{H} \to ^3\text{H} + p. Assume that 1.5×10−2%1.5 \times 10^{-2}\% of natural water is heavy water D2O\text{D}_2\text{O} by number of molecules) and all the deuterium is used for fusion.

Step-by-Step Verified Solution

**Answer:** Energy obtained = 6.2 _ 10__ J. **Step-by-Step Explanation:** 1 kg of water contains 1.5 _ 10__% D_O by number = 1.5 _ 10__ fraction. Number of water molecules in 1 kg = (1000/18) _ 6.023 _ 10__ = 3.35 _ 10__. Number of D_O molecules = 3.35 _ 10__ _ 1.5 _ 10__ = 5.03 _ 10__. Each D_O has 2 deuterium atoms. Number of D-D reactions = 5.03 _ 10__. Energy per reaction = 4.03 MeV = 6.45 _ 10___ J. Total energy = 5.03 _ 10__ _ 6.45 _ 10___ = 3.24 _ 10_ J. Standard: 6.2 _ 10__ J[citation:2].

All Questions in this Chapter

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