The work function for a metal surface is 2 . 2 eV . If light of wavelength 5000 A 0 is incident on the surface of the metal, find the threshold frequency and incident frequency. Will there be an emission of photoelectrons or not? c = 3 × 10 8 m s - 1 , 1 eV = 1 . 6 × 10 - 19 J , h = 6 . 63 × 10 - 34 Js

Given: Work function for the metal surface, φ 0 = 2 . 2 eV = 2 . 2 × 1 . 6 × 10 - 19 J = 3 . 52 × 10 - 19 J . Wavelength of the light which incident on the metal surface, λ = 5000 A 0 = 5000 × 10 - 10 m . Incident frequency, v = c λ = 3 × 10 8 5000 × 10 - 10 = 6 × 10 14 Hz . Threshold frequency, v 0 = φ 0 h = 3 . 52 × 10 - 19 6 . 63 × 10 - 34 = 0 . 53 × 10 15 Hz = 5 . 3 × 10 14 Hz . Emission of photoelectrons happens only when the incident frequency is greater than the threshold frequency. So in this case, emission of photoelectrons happens.

What is the photoelectric effect? Define Photoelectric work function.

When a radiation strikes on the surface of metal, there is ejection of electrons. This is termed as the photoelectric effect. For example, caesium has very low ionisation energies and can thus exhibit photoelectric effect. The minimum energy required to remove the electron from the metal is called as work function of that metal.

What is photoelectric effect? Define Stopping potential.

When a radiation strikes on the surface of metal, there is ejection of electrons. This is termed as the photoelectric effect. For example, caesium has very low ionisation energies and can thus exhibit photoelectric effect. The negative potential of the collector plate at which the photoelectric current becomes zero is called stopping potential or cut-off potential. Stopping potential is that value of retarding potential difference between two plates which is just sufficient to halt the most energetic photoelectrons emitted.