Question

(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 × 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?

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Solution

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

Specific
charge of an electron, *e/m*
= 1.76 ×
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

**(b)**Potential
of the anode, *V*
= 10 MV = 10 ×
10^{6} V

The speed of each electron is given as:

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
*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*
= *mc*^{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}

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