## anonymous 5 years ago Light falls on a photoelectric surface that has a work function of 2.90 eV. If the voltage required to stop the ejected electrons is 0.210 V, what is the wavelength of the incident radiation?

1. anonymous

$\theta=hf$ is the formula to solve this problem

2. anonymous

since work function is 2.90eV, nd Stopping Potential is0.210eV, so energy incident=Work function+ Max kinetic Energy(Stopping Potential), $h \nu=\phi+Kmax$ Therefore .$h \nu=$2.9eV+0.21eV =3.11eV=$hc/\lambda$, calculate $\lambda$=$hc\3.11\times1.6\times10^{-19}$=399.48nm (by calculation)