Answer
(a) $3.11\times 10^8m/s$
(b) $2.28\times 10^8m/s$
Work Step by Step
(a) We can find the final speed classically as
$\frac{1}{2}mv^2=eV$
This simplifies to:
$v=\sqrt{\frac{2eV}{m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{2(1.60\times 10^{-19})(276,000)}{9.11\times 10^{-31}}}$
$v=3.11\times 10^8m/s$
(b) We can find the final speed relativistically as
$K.E=m_{\circ}c^2(\frac{1}{\sqrt{1-v^2/c^2}}-1)=eV$
This simplifies to:
$v=c\sqrt{1-(1+\frac{eV}{m_{\circ}c^2})^{-2}}$
We plug in the known values to obtain:
$v=(3.00\times 10^8)\sqrt{1-(1+\frac{276KeV}{511KeV})^{-2}}$
$v=2.28\times 10^8m/s$
