#### Question

An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10^{−2} mm of Hg). A magnetic field of 2.83 × 10^{−4} T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method. Determine *e*/*m* from the data.

#### Solution

Potential of an anode, *V *= 100 V

Magnetic field experienced by the electrons,* B* = 2.83 × 10^{−4} T

Radius of the circular orbit *r* = 12.0 cm = 12.0 × 10^{−2} m

Mass of each electron = *m*

Charge on each electron = *e*

Velocity of each electron = *v*

The energy of each electron is equal to its kinetic energy, i.e.,

`1/2 mv^2 = eV`

`v^2 = (2eV)/m` ...(1)

It is the magnetic field, due to its bending nature, that provides the centripetal force

`(F = mv^2/r)` for the beam. Hence, we can write:

Centripetal force = Magnetic force

`mv^2/r = evB`

`eB = mv/r`

`v = (eBr)/m` ...(2)

Putting the value of *v* in equation (1), we get:

`"2eV"/m = (e^2B^2r^2)/m^2`

`e/m = "2V"/(B^2r^2)`

`= (2xx100)/((2.83xx 10^(-4))^2 xx (12 xx 10^(-2))^2)= 1.73 xx 10^11 C kg^(-1)`

Therefore, the specific charge ratio (*e/m*) is `1.73 xx 10^11 C Kg^(-1)`