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
- A1.5 10⁵ ~ms ⁻¹
- B2.5 10⁶ ~ms ⁻¹
- C2.1 10⁷ ~ms ⁻¹
- 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