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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
The Special Theory of Relativity

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Special Theory of Relativity

Calculate the speed of an electron whose kinetic energy is 10 keV .

Calculate the speed of an electron whose kinetic energy is 10 keV10 \text{ keV}.

Options

  1. A8.27×107 m s−18.27 \times 10^{7} \text{ m s}^{-1}
  2. B2.92×107 m s−12.92 \times 10^{7} \text{ m s}^{-1}
  3. C5.85×107 m s−15.85 \times 10^{7} \text{ m s}^{-1}
  4. D5.93×107 m s−15.93 \times 10^{7} \text{ m s}^{-1}

Correct answer

D. 5.93×107 m s−15.93 \times 10^{7} \text{ m s}^{-1}

Step-by-step solution

Given:
Kinetic energy of the electron, K=10 keV=10×103 eVK = 10 \text{ keV} = 10 \times 10^3 \text{ eV}

Converting kinetic energy into Joules:
K=104×1.6×10−19 J=1.6×10−15 JK = 10^4 \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-15} \text{ J}

Mass of an electron, m=9.1×10−31 kgm = 9.1 \times 10^{-31} \text{ kg}

Using the non-relativistic formula for kinetic energy:
K=12mv2K = \dfrac{1}{2} m v^2

Rearranging for speed vv:
v=2Kmv = \sqrt{\dfrac{2K}{m}}

Substituting the values:
v=2×1.6×10−159.1×10−31v = \sqrt{\dfrac{2 \times 1.6 \times 10^{-15}}{9.1 \times 10^{-31}}}

v=3.2×10−159.1×10−31v = \sqrt{\dfrac{3.2 \times 10^{-15}}{9.1 \times 10^{-31}}}

v=0.3516×1016v = \sqrt{0.3516 \times 10^{16}}

v≈5.93×107 m s−1v \approx 5.93 \times 10^7 \text{ m s}^{-1}

Answer: 5.93×107 m s−15.93 \times 10^{7} \text{ m s}^{-1}
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