NTA Abhyas JEE Main2020PhysicsDual Nature of MatterPractice
Light of wavelength 4000 Å is allowed to fall on a metal surface having work function 2 eV . The maximum velocity of the emitted electrons is ( h = 6.6 × 10 - 34 Js )
Options
- A1.35 × 10 5 m s - 1
- B2.7 × 10 5 m s - 1
- C6.2 × 10 5 m s - 1
- D8.1 × 10 5 m s - 1
Correct answer
C. 6.2 × 10 5 m s - 1
Step-by-step solution
1 2 m v 2 = h c λ - ϕ ( i n e V ) = 6.6 × 10 - 34 × 3 × 10 8 4000 × 10 - 10 × 1.6 × 10 - 19 - 2 = 3.1 - 2 = 1.1 e V = 1.1 × 1.6 × 10 - 19 J = 1.76 × 10 - 19 J v = 1.76 × 10 - 19 × 2 9 × 10 -3 1 = 6.2 × 10 5 m s - 1