Q20 of 43 Page 409

(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter.

Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg–1.


(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Given:


Potential difference between collector and emitter = 500V


Specific charge of electron (charge per unit mass e/m) = 1.76 × 1011 C


Kinetic energy of an electron is given by:


…(1)


Where,


M = mass of electron


v = velocity of electron


e = charge of electron


V = potential difference (accelerating potential)


(a) From equation (1), we can write


…(2)


By putting the values in equation (2) we can find electron velocity.



v = 1.327 × 107 ms-1


(b) Accelerating potential, V = 10MV = 106V


Let speed of electron be v1


Again putting the values in equation (2),


v1 =


v1 = 1.8 × 109ms-1


This result is wrong as we understand that speed of light


(i.e. 3 × 108 ms-1) is the theoretical limit of the speed.


Such problems can be dealt using relativistic mechanics,


Relativistic mass is given by:


m =


Where,


m = relativistic mass


m0 = rest mass


v = velocity of particle


c = speed of light


At relativistic speeds, kinetic energy is given by,


KE = mc2-m0c2


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Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

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What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

 21

(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s–1 is subject to a magnetic field of 1.30 × 10–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011C kg–1.

(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?


[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasize the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

 22

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.