User talk:HosseinJavanMard

Problem:

Radiation of wavelength $\lambda$ = 290 nm falls on a metal surface for which the work function is W = 4.05 eV. What potential is needed to stop the most energetic photoelectrons?

SOLUTION:

The energy of the photons is

$E=h\nu ={\frac {hc}{\lambda }}=$(6.63 X $10^{-34}$ J.s)(3 X $10^{8}$ m/s)/2.9 x $10^{-7}$ m) = 6.86 X $10^{-19}$ J = (6.86 X $10^{-19}$ J)/(1.60 X $10^{-19}$ J/eV) = 4.28 eV

The maximum kinetic energy of the photoelectron is $E-W=0.23eV$, so the potential difference is 0.23 volts.