Light of wavelength `6000Å` is incident on a metal. To release an electron from the metal surface, 1.77 eV of energy is needed. Find the kinetic energy of the fastest photoelectron. What is the threshold frequency of the metal? `(h = 6.62 xx 10^(-27) " erg".s, 1eV = 1.6 xx 10^(-12) " erg")`

Explanation

Energy of a photon,
`hf = h (c)/(lamda) = (6.62 xx 10^(-27) xx 3 xx 10^(10))/(6000 xx 10^(-8))`erg
`= (6.62 xx 10^(-27) xx 3 xx 10^(10))/(6000 xx 10^(-8) xx 1.6 xx 10^(-12)) eV = 2.07 eV`
As per Einstein's photoelectric equation
`E_("max") = hf - W_(0) = 2.07 - 1.77 = 0.3eV`
Threshold frequency,
`f_(0) = (W_(0))/(h) = (1.77 xx 1.6 xx 10^(-12))/(6.62 xx 10^(-27)) = 4.28 xx 10^(14) Hz`