Consider a metal exposed to light of wavelength $600 \, nm$. The maximum kinetic energy of the emitted electron doubles when light of wavelength $400 \, nm$ is used. Find the work function of the metal in $eV$. (in $, eV$)

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the metallic surface is:

$A$ photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_0$. The expression for the de-Broglie wavelength $\lambda_e$ is ($m = \text{mass of electron}$, $h = \text{Planck's constant}$, $c = \text{speed of light}$):

The work function of a surface of a photosensitive material is $6.2\, eV$. The wavelength of the incident radiation for which the stopping potential is $5\, V$ lies in the

If a metal sheet is irradiated with radiations of frequencies $v_1$ and $v_2$ and the kinetic energies of the emitted photoelectrons are in the ratio $1 : x$, then the threshold frequency of the metal is:

The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is