(a) Estimate the speed with which electrons emitted from a heated emitter

Solution:

ANSWER:

 

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

Specific charge of an electron, e/m = 1.76 × 1011 C kg−1

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

$\mathrm{KE}=\frac{1}{2} m v^{2}=e V$

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

$=\left(2 \times 500 \times 1.76 \times 10^{11}\right)^{\frac{1}{2}}=1.327 \times 10^{7} \mathrm{~m} / \mathrm{s}$

Therefore, the speed of each emitted electron is $1.327 \times 10^{7} \mathrm{~m} / \mathrm{s}$.

(b)Potential of the anode, V = 10 MV = 10 × 106 V

The speed of each electron is given as:

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

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

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

This result is wrong because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << 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 = mc2

Where,

m = Relativistic mass

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

m0 = Mass of the particle at rest

Kinetic energy is given as:

 

K = mc2 − m0c2