Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.20
NCERT Solutions for Class 12 Physics Chapter 11 Question-20
Additional Exercise
(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?
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Ans (a).
Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 x 1011 C kg-1
The speed of each emitted electron is given by the relation for kinetic energy as:
KE = 1/2 mv²=eV
Therefore ,v =√(2eV/m) = √(2V e/m)
= √(2x 500 x 1.76 x 10¹¹) =1.327 x 107 m/s
Therefore, the speed of each emitted electron is 1.327 x 107 m/s.
Ans (b).
Potential of the anode, V = 10 MV = 10 x 106 V
The speed of each electron is given as:
v = √ (2V e/m))
= √ (2 x 107 x 1.76x 10¹¹)- = 1.88 x 109 m/s
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 = me2
Where, m = Relativistic mass = mo √ (1 – v2/c2)
mo = Mass of the particle at rest Kinetic energy is given as: K = mc2 – moc2