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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

Find the speed of an electron that has a kinetic energy of 10 MeV .

Find the speed of an electron that has a kinetic energy of

10 \text{ MeV}

.

Options

A $2.996 \times 10^{8} \text{ m s}^{-1}$
B $1.500 \times 10^{8} \text{ m s}^{-1}$
C $3.000 \times 10^{8} \text{ m s}^{-1}$
D $2.950 \times 10^{8} \text{ m s}^{-1}$

Correct answer

A. $2.996 \times 10^{8} \text{ m s}^{-1}$

Step-by-step solution

The rest mass energy of an electron is

E_0 = m_0 c^2 = 0.511 \text{ MeV}

.

The kinetic energy is

K = 10 \text{ MeV}

.

The total energy of the electron is given by

E = K + E_0 = 10 + 0.511 = 10.511 \text{ MeV}

.

Using the relativistic energy relation

E = \dfrac{m_0 c^2}{\sqrt{1 - \dfrac{v^2}{c^2}}}

, we get:

10.511 = \dfrac{0.511}{\sqrt{1 - \dfrac{v^2}{c^2}}}

\sqrt{1 - \dfrac{v^2}{c^2}} = \dfrac{0.511}{10.511} \approx 0.0486

Squaring both sides:

1 - \dfrac{v^2}{c^2} \approx (0.0486)^2 \approx 0.00236

\dfrac{v^2}{c^2} \approx 1 - 0.00236 = 0.99764

\dfrac{v}{c} \approx \sqrt{0.99764} \approx 0.9988

Taking

c = 3 \times 10^8 \text{ m s}^{-1}

:

v \approx 0.9988 \times 3 \times 10^8 \text{ m s}^{-1} = 2.996 \times 10^8 \text{ m s}^{-1}

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