Question

Light of wavelength 2000 Å falls on an aluminium surface. In aluminium, 4.2 eV are required to remove an electron. Determine (i) KE of the fastest emitted photo-electron (ii) KE of the slowest emitted photo electron (iii)stopping potential and (iv) cut-off wavelength for aluminium.

Answer 1

Photon energy of the incident light is

= 2 eV = 2 × 1.6 × 10^–19 = 3.2 × 10^–19 J