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Exercises Relativity, quantum and nuclear

Quick check from College Physics, 28.6 Relativistic Energy

Relativistic Energy: quick check

A photon decays into an electron-positron pair. What is the kinetic energy of the electron if its speed is $0.992 c$ ?

Answer it in your own words before you look.

Answer
$\begin{array}{l} {KE}_{rel} & = & (γ - 1) {mc}^{2} = (\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} - 1) {mc}^{2} \\ = & (\frac{1}{\sqrt{1 - \frac{(0.992 c)^{2}}{c^{2}}}} - 1) (9.11 \times 10^{- 31} kg) (3.00 \times 10^{8} m/s)^{2} = 5.67 \times 10^{- 13} J \end{array}$
28.64

How did it go?

PreviousComparing Kinetic Energy: Relativistic Energy Versus Classical Kinetic Energy Next Calculating Photon Energy and the Photoelectric Effect: A Violet Light

From College Physics (1st edition) by OpenStax (Rice University), section 28.6, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.