Question

Draw a neat, labelled energy level diagram for H atom showing the transitions. Explain the series of spectral lines for H atom, whose fixed inner orbit numbers are $3$ and $4$ respectively. The work functions for potassium and caesium are $2.25$eV and $2.14$eV respectively. Is the photoelectric effect possible for either of them if the incident wavelength is $5180\stackrel{o}{A}$?
[Given: Planck's constant$=6.63\times {10}^{-34}$J.s;
Velocity of light $=3\times {10}^{8}m/s;1eV=1.6\times {10}^{-19}$J].

Answer 1

Photoelectric Effect:

Incident wavelength $\lambda =5180\times {10}^{-10}m$
Energy, $E=\frac{hc}{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{5180\times {10}^{-10}}=3.84\times {10}^{-19}J=2.38eV$
Equation for Photoelectric Effect : $\frac{hc}{\lambda }-\Phi =K.E$
where $\frac{hc}{\lambda }$ is energy of incident photos, $|Phi$ is work function of material and $K.E$ is kinetic energy of emitted electron.
Now, for both Potassium and Cesium the work function is less than the energy of incident photons which means that the kinetic energy of the emitted electrons will be positive. Therefore, the photoelectric effect is possible.