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?

Solution

Here,
Potential, V = 10 MV = $10 \times 10^{6} = 10^{7} V$
Specific charge ratio, $\frac{e}{m} = 1.76 \times 10^{11} C {kg}^{- 1}$

Now,
$v = \sqrt{\frac{2 eV}{m}} = \sqrt{2 \times 1.76 \times 10^{11} \times 10^{7}}$
i.e., $v = 1.876 \times 10^{9} {ms}^{- 1}$

The obtained speed is not possible because, it is greater than the speed of light. Now, since, $[m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}]$ therefore, as v approaches c, mass of the electron becomes infinite.
Hence, the formula needs to be modified.

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Wave Optics

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?

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