When a photon of energy $3.8 \,eV$ falls on a metallic surface of work function $2.8 \,eV$,then the kinetic energy of the emitted electrons is .......... $eV$.

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

Explain: "For photoelectric emission, the frequency of incident radiation should be greater than the threshold frequency."

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

The stopping potential $V_0$ for photoelectric emission from a metal surface is plotted along the $Y-$ axis and the frequency $\nu$ of incident light along the $X-$ axis. $A$ straight line is obtained as shown. Planck's constant $h$ is given by: