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QPage 458: Q26: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q26 Solution: Find the speed of an electron with kinetic energy a) 1 eV, b)…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply the classical formula $K = \frac{1}{2}mv^2$ to all parts, which yields unphysical speeds exceeding $c$ for high energies like 10 MeV. Always use the relativistic relation $K = (\gamma - 1)m_0c^2$ to ensure $v < c$.

Find the speed of an electron with kinetic energy a) 1 eV, b) 10 keV and c) 10 MeV.

Step-by-Step Verified Solution

**Answer:** (a) v=5.9×105v = 5.9 \times 10^5 m/s (b) v=5.8×107v = 5.8 \times 10^7 m/s (c) v=2.996×108v = 2.996 \times 10^8 m/s. **Step-by-Step Explanation:** Using K=(γ−1)m0c2K = (\gamma - 1)m_0c^2: (a) K=1K = 1 eV = 1.6×10−191.6 \times 10^{-19} J. γ−1=1.6×10−19/(9.1×10−31×9×1016)=1.95×10−6\gamma - 1 = 1.6 \times 10^{-19}/(9.1 \times 10^{-31} \times 9 \times 10^{16}) = 1.95 \times 10^{-6}. v≈c2(γ−1)=3×108×3.9×10−6=5.9×105v \approx c\sqrt{2(\gamma-1)} = 3 \times 10^8 \times \sqrt{3.9 \times 10^{-6}} = 5.9 \times 10^5 m/s. (b) K=10K = 10 keV: v=5.8×107v = 5.8 \times 10^7 m/s. (c) K=10K = 10 MeV: v=2.996×108v = 2.996 \times 10^8 m/s .

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