JEE Main20199 Apr 2019Morning ShiftPhysicsDual Nature of MatterActual
The electric field of light wave is given as E → = 10 – 3 cos ⁡ 2 π x 5 × 10 – 7 - 2 π × 6 × 10 14 t x ^ N C . This light falls on a metal plate of work function 2 e V . The stopping potential of the photo-electrons is: Given, E (in e V ) = 12375 λ ( i n Å )
Options
- A0.72 V
- B2.0 V
- C2.48 V
- D0.48 V
Correct answer
D. 0.48 V
Step-by-step solution
E → = 10 - 3 c o s 2 π x 5 × 10 - 7 - 2 π × 6 × 10 14 t x ^ N / c k = 2 π 5 × 10 - 7 And ω = 6 × 10 14 × 2 π ⇒ f = 6 × 10 14 H z E = ϕ + K E m a x K E m a x = h f - ϕ = 6.6 × 10 - 34 × 6 × 10 14 1.6 × 10 - 19 - 2 = 2.475 - 2 = 0.48 e V ∴ Stopping potential is 0.48 V