$A$ photosensitive surface made of cesium on tungsten is irradiated by monochromatic radiation of wavelength $640.2 \ nm$ $(1 \ nm = 10^{-9} \ m)$ from a neon bulb. The measured stopping potential is $0.54 \ V$. If the source is replaced by another source and the same photocell is irradiated by a line of $427.2 \ nm$,what will be the new stopping potential in $V$?

In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

In the graph of maximum kinetic energy of photoelectrons emitted from a metal surface versus the frequency of incident radiation based on Einstein's photoelectric equation,the slope is .......

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

When a metal 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 $V/3$. The threshold wavelength for the surface is:

$(i)$ In the explanation of the photoelectric effect,we assume one photon of frequency $f$ collides with an electron and transfers its energy. This leads to the equation for the maximum kinetic energy $E_{max}$ of the emitted electron as $E_{max} = hf - \phi_0$ (where $\phi_0$ is the work function of the metal). If an electron absorbs $2$ photons (each of frequency $f$),what will be the maximum energy for the emitted electron?
$(ii)$ Why is this fact (two-photon absorption) not taken into consideration in our discussion of the stopping potential?