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QPage 458: Q25: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q25 Solution: Through what potential difference should an electron be…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug $v=0.6c,0.9c,0.99c$ into the non‑relativistic formula $K=\frac12mv^2$, which severely underestimates the required voltage. Always use the relativistic kinetic energy $K=(\gamma-1)m_0c^2$ with $\gamma=1/\sqrt{1-v^2/c^2}$ before equating to $eV$.

Through what potential difference should an electron be accelerated to give it a speed of a) 0.6c, b) 0.9c and c) 0.99c?

Step-by-Step Verified Solution

**Answer:** (a) V=0.128V = 0.128 MV (b) V=0.66V = 0.66 MV (c) V=3.1V = 3.1 MV. **Step-by-Step Explanation:** K=(γ−1)m0c2=eVK = (\gamma - 1)m_0c^2 = eV. (a) v=0.6cv = 0.6c: γ=1.25\gamma = 1.25. K=0.25×0.511=0.128K = 0.25 \times 0.511 = 0.128 MeV. V=0.128V = 0.128 MV. (b) v=0.9cv = 0.9c: γ=2.294\gamma = 2.294. K=1.294×0.511=0.66K = 1.294 \times 0.511 = 0.66 MeV. V=0.66V = 0.66 MV. (c) v=0.99cv = 0.99c: γ=7.089\gamma = 7.089. K=6.089×0.511=3.11K = 6.089 \times 0.511 = 3.11 MeV. V=3.1V = 3.1 MV.

All Questions in this Chapter

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