When a photon of energy $hv$ is incident on an aluminum plate (work function $E_0$),photoelectrons with maximum kinetic energy $K$ are emitted. What will be the maximum kinetic energy of the emitted photoelectrons when a photon of energy $2hv$ is incident on the same aluminum plate?

The threshold frequency for a metallic surface corresponds to an energy of $6.2 \ eV$ and the stopping potential for a radiation incident on this surface is $5 \ V$. The incident radiation lies in:

The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda$ (where $\lambda < \lambda_{0}$) will be:

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :

Assertion : In the process of photoelectric emission, all emitted electrons do not have the same kinetic energy.
Reason : If radiation falling on the photosensitive surface of a metal consists of different wavelengths, then the energy acquired by electrons absorbing photons of different wavelengths shall be different.

The stopping potential in the context of the photoelectric effect depends on the following property of incident electromagnetic radiation: