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 × 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?

Answer 1

(a) anode maintained at a potential difference , V = 500Volts.

specific charge, e/m = 1.76 × 10¹¹ C/kg....(1)

kinetic energy = energy of accelerated electron

or, 1/2 mv² = eV

or, v = √{2Ve/m}

= √{2V(e/m)}

from equation (1),

v = √{ 2 × 500 × 1.76 × 10¹¹} = 1.33 × 10^7 m/s

(b) for anode potential, V = 10MV = 10^7 Volts

speed of electron, v = √{2V(e/m)}

= √{2 × 10^7 × 1.76 × 10¹¹}

= 1.88 × 10^9 m/s > 3 × 10^8 m/s ( speed of light in vaccum)

so, 1.88 × 10^9 m/s is impossible . since nothing can move with the speed grater than the speed of light in vaccum.

The formula for kinetic energy , E = 1/2 mv² is valid only for v << c . for the situation when v is comparable to speed of light c, we use relativistic formula,

The relativistic mass is given by,

$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ where $m_0$ is the rest mass .

also total energy is given by,

$T.E^2=c^2p^2+(m_0c^2)^2$

= $c^2(p^2+m_0^2c^2)$

putting, $p=mv=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$

so, $T.E^2=c^2\left[\frac{m_0^2v^2}{1-\frac{v^2}{c^2}}+m_0^2c^2\right]$

$T.E^2=\frac{m_0^2c^4}{1-\frac{v^2}{c^2}}$

or, $T.E=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}$

now, Total energy = kinetic energy + $m_0c^2$

so, kinetic energy = T.E - $m_0c^2$

or, eV = $\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2$

or, $\frac{eV}{m_0c^2}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1$

or, $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=1+\frac{1.6\times10^{-19}\times10^7}{9.1\times10^{-31}\times(3\times10^{-8})^{-2}}$

or, $19.536+1=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

or, $1-\frac{v^2}{c^2}=0.00237$

or, $\frac{v^2}{c^2}=0.997$

or, speed , v = √(0.997)c = 0.999c