$(a)$ Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500\;V$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron,$i.e.$,the $e/m$ is given to be $1.76 \times 10^{11}\; C\; kg^{-1}$.
$(b)$ Use the same formula you employ in $(a)$ to obtain electron speed for a collector potential of $10\;MV$. Do you see what is wrong? In what way is the formula to be modified?

(A) Potential difference $V = 500\;V$. Specific charge $e/m = 1.76 \times 10^{11}\;C\;kg^{-1}$.
Using the conservation of energy,$eV = \frac{1}{2}mv^2$,we get $v = \sqrt{2V(e/m)}$.
$v = \sqrt{2 \times 500 \times 1.76 \times 10^{11}} = \sqrt{1.76 \times 10^{14}} \approx 1.33 \times 10^7\;m/s$.
$(b)$ For $V = 10\;MV = 10^7\;V$,the formula gives $v = \sqrt{2 \times 10^7 \times 1.76 \times 10^{11}} = \sqrt{3.52 \times 10^{18}} \approx 1.88 \times 10^9\;m/s$.
This result is wrong because $v > c$ (speed of light),which is physically impossible.
The formula $K.E. = \frac{1}{2}mv^2$ is non-relativistic. For high potentials,we must use the relativistic kinetic energy formula: $K.E. = (\gamma - 1)mc^2$,where $\gamma = (1 - v^2/c^2)^{-1/2}$.