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(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?
Asked by Yashraj Last Modified
Answer 
Harshit Bhardwaj
(a)Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 × 1011 C kg−1
The speed of each emitted electron is given by the relation for kinetic energy as:
Therefore, the speed of each emitted electron is 
(b)Potential of the anode, V = 10 MV = 10 × 106 V
The speed of each electron is given as:
This result is wrong because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v << c.
For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as:
E = mc2
Where,
m = Relativistic mass
m0 = Mass of the particle at rest
Kinetic energy is given as:
K = mc2 − m0c2
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