Ncert Exercises & Solutions

Question 11.20:(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., its e/ m is given to be $1.76 x 10^{11} C {kg}^{- 1}$ .

(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)
Hint: $\mathrm{v}=\left(\frac{2 \mathrm{eV}}{\mathrm{m}}\right)^{\frac{1}{2}}$
Step 1: Find the velocity of the emitted electron.
The potential difference across the evacuated tube, V = 500 V
The specific charge of an electron. e/m =

1.76 x 10^{11} C {kg}^{- 1}

The kinetic energy of the electron is given by:

\begin{array}{l} KE = \frac{1}{2} {mv}^{2} = eV \\ ∴ v = {(\frac{2 eV}{m})}^{\frac{1}{2}} = {(2 V \times \frac{e}{m})}^{\frac{1}{2}} \\ = {(2 \times 500 \times 1 .76 \times 10^{11})}^{\frac{1}{2}} = 1 .327 \times 10^{7} m / s \end{array}

Therefore, the speed of each emitted electron is

1.327 x 10^{7} m / s

(b)
Hint: $\mathrm{v}=\left(\frac{2 \mathrm{eV}}{\mathrm{m}}\right)^{\frac{1}{2}}$

Step 1: Find the velocity of the emitted electron.
The speed of each electron is given as:

\begin{aligned} v & = {(2 V \frac{e}{m})}^{\frac{1}{2}} \\ = {(2 \times 10^{7} \times 1 .76 \times 10^{11})}^{\frac{1}{2}} \\ = 1 .88 \times 10^{9} m / s \end{aligned}

Step 2: Identify the mistake in the above answer.
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} {(1 - \frac{v^{2}}{c^{2}})}^{\frac{-1}{2}}

m =Mass of the particle at rest.

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