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(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 500V 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 10MV 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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