A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4 . 3 × 10 14 Hz . The velocity of ejected electron is _ _ _ × 10 5 ms - 1 (Nearest integer) Use : h = 6 . 63 × 10 - 34 Js , m e = 9 . 0 × 10 - 31 kg

Step-by-step solution

λ = 500   nm → v = C λ = 3 × 10 8 500 × 10 - 9 = 6 × 10 14   Hz v 0 = 4 . 3 × 10 14   Hz For photo electric effect, hv = hv 0 + KE KE = hv - hv 0 = h v - v 0 = 6 . 6 × 10 - 34 6 × 10 14 - 4 . 3 × 10 14 = 6 . 6 × 1 . 7 × 10 - 20   J   K . E .  = 1 2 mv 2 ; v = 2 × K · E m V = 2 × 6 . 6 × 1 . 7 × 10 - 20 9 . 1 × 10 - 31 = 24 . 65 × 10 10 = 5 × 10 5   m / s