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

Numerical

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 × 1011 C kg−1

Therefore, the specific charge ratio (e/m) is 1.73 × 1011 C kg−1.

Concept: Photoelectric Effect and Wave Theory of Light
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Chapter 11: Dual Nature of Radiation and Matter - Exercise [Page 409]

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NCERT Physics Class 12
Chapter 11 Dual Nature of Radiation and Matter
Exercise | Q 11.22 | Page 409
NCERT Physics Class 12
Chapter 11 Dual Nature of Radiation and Matter
Exercise | Q 22 | Page 409
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