Light of wavelength `5000 Å` falls on a sensitive plate with photoelectric work function of `1.9 eV`. The kinetic energy of the photoelectron emitted will be

Options

A. 0.58 eV
B. 2.48 eV
C. 1.24 eV
D. 1.16 eV

Answer

D

Explanation

Energy of photon is given by
`E = (h c)/(lambda) = (12375)/(lambda (Å)) eV`
`therefore E = (12375)/(5000) = 2.48 eV`
Einstein.s photoelectric equation is
`E_(k) = E-w = 2.48 eV - 1.9 eV = 0.58 eV`