(a) Estimate the speed with which electrons emitted from a heatedemitter of an evacuated tube impinge on the collector maintainedat a potential difference of 500 V with respect to the emitter.Ignore the small initial speeds of the electrons. Thespecific charge of the electron, i.e., its e/m is given to be1.76 × 1011 C kg–1. (b) Use the same formula you employ in (a) to obtain electron speedfor an collector potential of 10 MV. Do you see what is wrong ? Inwhat way is the formula to be modified?

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Solution

a)

Given: The potential difference between collector and emitter is 500 V and the specific charge of the electron is 1.76× 10 11 C/kg.

The work done on the electron by the potential difference is equal to the kinetic energy of the electron.

1 2 m v 2 =eV v= e m ×2V

Where, v is the speed of electron, e m is the specific charge of electron and V is the potential difference.

By substituting the values in the above equation, we get

v= ( 1.76× 10 11 )×2×500 =1.33× 10 7 m/s

Thus, the speed of the electron is 1.33× 10 7 m/s.

b)

The speed of the electron for potential difference 10 MV is,

v= ( 1.76× 10 11 )×2×10× 10 6 =1.876× 10 9 m/s

This value of speed is not possible because it is greater than the speed of light. The above formula is valid only for speed v<<c.

When the speed of particle approaches to the speed of light, the mass of the particle starts to change according to the relation,

m= m 0 1− v 2 c 2

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Similar questions

(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 an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

Q. Choose the correct answer from the alternatives given.
Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The specific charge of the electron i.e., its elm is given to be

1.76 \times 10^{11} C k g^{- 1}

(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 × 10¹¹ C kg⁻¹.

(b) Is the formula you employ in (a) valid for calculating 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 emphasise 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.]

Q. An electron beam accelerated from rest through a potential difference of

500 V

in vacuum is allowed to impinge on a surface normally. The incident current is

50 μ A

and if the electrons come to rest on striking the surface the force on it is

Q. In an X-ray tube, electrons accelerated through a potential difference of 15000 V strike a copper target. The speed of the emitted X-rays inside the tube is
[e= charge on electron, m=mass of electron, Z= atomic number of target]

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