Question

Write Einstein's photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation.
The maximum kinetic energy of the photo electrons gets doubled when the wavelength of light incident on the surface changes from ${\lambda }_1$ to ${\lambda }_2$. Derive the expressions for the threshold wavelength ${\lambda }_0$ and work function for the metal surface.

Answer 1

Einstein's photoelectric equation is given as

$h\nu ={\varphi }_0+K_{max}$

$h\nu =$Energy of photon

${\varphi }_0$= Work function of metal

$K_{max}$=Maximum kinetic energy of emitted photoelectron

Explanation of features of photoelectric effect-

(a) Einstein said that one photoelectron is ejected from a metal surface if one photon of suitable light radiation falls on it. If the intensity of the light is increased, the number of incident photon increases, which results in an increase in the number of photo-electrons ejected. This implies photo current is proportional to intensity and radiation.

(b) If $\nu <{\nu }_0$ than maximum K.E. is negative, which is impossible. Hence photoelectric emission does not take place for the incident radiation below threshold frequency.

(c) If $\nu >{\nu }_0$, maximum K.E. increases. This means, maximum K.E. of photoelectrons depends only the frequency of incident light.

Maximum kinetic energy of photoelectron

$=K_{max}=\frac{hc}{{\lambda }_1}-{\varphi }_0$....(i)

and $2K_{max}=\frac{hc}{{\lambda }_2}-{\varphi }_0$

$⟹K_{max}=\frac{hc}{2{\lambda }_2}-\frac{{\varphi }_0}{2}$......(ii)

From (i) and (ii),

${\varphi }_0=hc(\frac{2}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

$⟹\frac{hc}{{\lambda }_0}=hc(\frac{2}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

where ${\lambda }_0$ is the threshold wavelength.

Thus ${\lambda }_0=\frac{{\lambda }_1{\lambda }_2}{2{\lambda }_2-{\lambda }_1}$