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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) Determine E/M From the Data. - Physics

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.

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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.,

`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)`

Concept: Photoelectric Effect and Wave Theory of Light
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NCERT Class 12 Physics Textbook
Chapter 11 Dual Nature of Radiation and Matter
Q 22 | Page 409
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