Answer to Question #175067 in Quantum Mechanics for AHSAN UL HAQ

Question

The kinetic energy of an electron is equal to its rest mass energy. Determine the

magnitude of its velocity and momentum.

Answer 1

$E_0=m_0c^2=9.1\cdot10^{-31}\cdot(3\cdot10^{8})^2=819\cdot10^{-16}\ (J)\approx512\ (keV)$

$KE=m_0c^2\bigg(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\bigg)$ , $KE=E_0$

$m_0c^2=m_0c^2\bigg(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\bigg)\to 1=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1$

$\sqrt{1-\frac{v^2}{c^2}}=\frac{1}{2}\to1-\frac{v^2}{c^2}=\frac{1}{4}\to v=\frac{\sqrt3}{2}c$ . Answer

Total energy

$E=m_0c^2/\sqrt{1-\frac{v^2}{c^2}}=512/\sqrt{1-\frac{(\frac{\sqrt3}{2}c)^2}{c^2}}=1024\ (keV)$

Momentum

$E^2=p^2c^2+(m_0c^2)^2\to p=\sqrt{(E^2-(m_0c^2)^2)/c^2}=$

$=\sqrt{(1024^2-(512)^2)/c^2}=887\ (keV/c)$ . Answer