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KCET2020PhysicsDual Nature of Matter

A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 ~V . The speed of the electron when it strikes the anode is

Options

  1. A1.5 10⁵ ~ms ⁻¹
  2. B2.5 10⁶ ~ms ⁻¹
  3. C2.1 10⁷ ~ms ⁻¹
  4. D2.5 10⁸ ~ms ⁻¹

Correct answer

C. 2.1 10⁷ ~ms ⁻¹

Step-by-step solution

Given, anode potential, V=1200 ~V Electron will accelerate with the effect of anode potential. Hence, 1 2 m v²= eV aligned v &= 2 e V m = 2 1.6 10⁻¹⁹ 1200 9.1 10⁻³¹ &= 421.98 10¹² =20.5 10⁶ ~ms ⁻¹ &=2.05 10⁷ ~ms ⁻¹ 2.1 10⁷ ~ms ⁻¹ aligned

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