(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 × 10¹¹ 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?

Ans (a).

Potential difference across the evacuated tube, V = 500 V

Specific charge of an electron, e/m = 1.76 x 10¹¹ C kg^-1

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

KE = 1/2 mv²=eV

Therefore ,v =√(2eV/m) = √(2V e/m)

= √(2x 500 x 1.76 x 10¹¹) =1.327 x 10⁷ m/s

Therefore, the speed of each emitted electron is 1.327 x 10⁷ m/s.

Ans (b).

Potential of the anode, V = 10 MV = 10 x 10⁶ V

The speed of each electron is given as:

v = √ (2V e/m))

= √ (2 x 10⁷ x 1.76x 10¹¹)- = 1.88 x 10⁹ m/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 = me²

Where, m = Relativistic mass = mo √  (1 – v²/c²)

mo = Mass of the particle at rest Kinetic energy is given as: K = mc² – moc²