Question

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?

Answer 1

\[\theta=hf\] is the formula to solve this problem

Answer 2

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)