The metallic surface is illuminated with monochromatic light of wavelength $\lambda$ and the stopping potential for the photoelectric current is $5V_0$. When the same metallic surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. What is the threshold wavelength for the surface?

$A$ 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 light rays having photons of energy $4.2\,eV$ are falling on a metal surface having a work function of $2.2\,eV$. The stopping potential of the surface is $.........\,V$.

In case of photoelectric emission from a certain metal, the cutoff frequency is $\nu$. If radiation of frequency $3\nu$ is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ($m$ = mass of electron, $h$ = Planck's constant).

The threshold frequency of a photosensitive material is equal to the frequency of the $H_{\alpha}$ line of hydrogen. If a photon whose frequency is equal to the frequency of the $H_{\beta}$ line of hydrogen is incident on this photosensitive material,the maximum kinetic energy of the emitted photoelectrons is ($R$ = Rydberg's constant,$h$ = Planck's constant,and $c$ = speed of light in vacuum).

When a photosensitive surface is irradiated by light of wavelengths $\lambda_1$ and $\lambda_2$,the kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively. The work function of the photosensitive surface is: