    Question

# (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  Suggest Corrections  0      Explore more