An electron gun with its collector at a potential of 100 V fires
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, = 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.,

$\frac{1}{2} m v^{2}=e V$

$v^{2}=\frac{2 e V}{m}$     ….(1)

It is the magnetic field, due to its bending nature, that provides the centripetal force $\left(F=\frac{m v^{2}}{r}\right)$ for the beam. Hence, we can write:

Centripetal force = Magnetic force

$\frac{m v^{2}}{r}=e v B$

$e B=\frac{m v}{r}$

$v=\frac{e B r}{m}$     …(2)

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

$\frac{2 e V}{m}=\frac{e^{2} B^{2} r^{2}}{m^{2}}$

$\frac{e}{m}=\frac{2 V}{B^{2} r^{2}}$

$=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^{2} \times\left(12 \times 10^{-2}\right)^{2}}=1.73 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$

Therefore, the specific charge ratio $(e / m)$ is $1.73 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$.

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