Light of frequency 7.21 × 10 14 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10 5 m s - 1 are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Step-by-step solution

Given, frequency of light, ν = 7.21 × 10 14 Hz Mass of electron, m = 9.1 × 10 − 31 kg Maximum speed of electrons, v max = 6 × 10 5 m/s Let ν 0 be the threshold frequency. Use the formula for kinetic energy K E = 1 2 m v max 2 = h ν − h ν 0 i.e., 1 2 × 9.1 × 10 − 31 × 6 × 10 5 × 6 × 10 5 = 6.63 × 10 − 34 ( ν − ν 0 ) or ν - ν 0 = 3 6 × 9 . 1 × 1 0 - 2 1 2 × 6 . 6 3 × 1 0 - 3 4 = 2 . 4 7 × 1 0 1 4 or ν 0 = 7.21 × 10 14 − 2.47 × 10 14 ( ∵ ν = 7.21 × 10 14 Hz ) = 4.74 × 10 14 Hz .