Determine the speed at which the kinetic energy of a particle will deviate by 1\% from its nonrelativistic value (1)/(2) m_0 v^2 .

Question

Quantrex Academy · Accepted Answer

Correct answer: D. 3.46 × 10^7 m/s

The relativistic kinetic energy is given by

K = m_0 c^2 (\gamma - 1) = m_0 c^2 \left(

\dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} -

1 \right)

.

Using the binomial expansion for

v \ll c

, we get:

K = m_0 c² \left(1 + \dfrac{1}{2} \dfrac{v^2}{c^2} + \dfrac{3}{8} \dfrac{v^4}{c^4} + \dots -1 \right)

K \approx \dfrac{1}{2}m_0 v² + \dfrac{3}{8}m_0 \dfrac{v^4}{c^2}

The nonrelativistic kinetic energy is

K_{nr} = \dfrac{1}{2}m_0 v²

.

The fractional deviation is given by:

\dfrac{K - K_{nr}}{K_{nr}} = \dfrac{\dfrac{3}{8} m_0 \dfrac{v^4}{c^2}}{\dfrac{1}{2} m_0 v^2} = \dfrac{3}{4} \dfrac{v^2}{c^2}

Given that the deviation is

1\% = 0.01

:

\dfrac{3}{4} \dfrac{v^2}{c^2} =0.01

v^2 = \dfrac{0.04}{3}c^2

v = \dfrac{0.2}{\sqrt{3}}c

Substituting

c = 3 \times10^8 \text{ m/s}

:

v = \dfrac{0.2}{\sqrt{3}} \times3 \times10^8 = 0.2 \times \sqrt{3} \times10^8

v \approx0.2 \times1.732 \times10^8 = 3.464 \times10^7 \text{ m/s}

Determine the speed at which the kinetic energy of a particle will...