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 = (2"eV")/"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:

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

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

= `(2 xx 100)/((2.83 xx 10^(-4))^2 xx (12 xx 10^(-2))^2)`

= 1.73 × 10^{11} C kg^{−1}

Therefore, the specific charge ratio (e/m) is 1.73 × 10^{11} C kg^{−1.}