## University Physics with Modern Physics (14th Edition)

At such a high speed, the electrons are relativistic. a. $2.06 \times10^6 volts$. b. $2.06 \times10^6 eV=3.3\times10^{-13}J$.
a. $\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}=5.025$ The relativistic kinetic energy is $(\gamma-1)mc^2=4.025mc^2=3.3\times10^{-13}J$. Convert that energy to electron volts, $K=2.06 MeV$. The potential difference required is $2.06 \times10^6 volts$. b. From part A, $K = 3.3\times10^{-13}J = 2.06 MeV$.