$(a)$ 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 inttial speeds of the electrons. The specific charge of the electron, $i.e.$, the $e / m$ is glven to be $1.76 \times 10^{11}\; C\; kg ^{-1}$

$(b)$ Use the same formula you employ in $(a)$ to obtain electron speed for an collector potential of $10 \;MV$. Do you see what is wrong? In what way is the formula to be modified?

$(a)$ Potential difference across the evacuated tube, $V=500\, V$

Specific charge of an electron, $e / m$ $=1.76 \times 10^{11}\, C\,kg ^{-1}$

The speed of each emitted electron is given by the relation for kinetic energy as

Therefore, the speed of each emitted electron is $K E=\frac{1}{2} m v^{2}=e v$

$\therefore v=\left(\frac{2 e V}{m}\right)^{\frac{1}{2}}=\left(2 V \times \frac{e}{m}\right)^{\frac{1}{2}}$

$=\left(2 \times 500 \times 1.76 \times 10^{11}\right)^{\frac{1}{2}}=1.366 \times 10^{7}\, m / s$

$(b)$ Potential of the anode, $V=10 \,MV =10 \times 10^{6} \,V =10^{7}$

The speed of each electron is given as

$v=\left(2 V \frac{e}{m}\right)^{\frac{1}{2}}$

$=\left(2 \times 10^{7} \times 1.76 \times 10^{11}\right)^{\frac{1}{2}}$

$=1.88 \times 10^{9} \,m / s$

This result is wrong because nothing can move faster than light. In the above formula, the

expression $(mv^{2}/2)$ for energy can only be used in the non-relativistic limit, i.e., for $\mathrm { v } < < \mathrm { c }$

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:

$E = mc ^{2}=m_{0}\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{1}{2}} Where$

$m =$ Relativistic mass $m _{0}=$ Mass of the particle at rest Kinetic energy is given as:

$K = mc ^{2}- m _{0} c ^{2}$