Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$,the wavelength must be changed to $\lambda'$ where

The electron in the hydrogen atom jumps from excited state $(n = 3)$ to its ground state $(n = 1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \, eV$,the stopping potential is estimated to be (the energy of the electron in $n^{th}$ state $E_n = \frac{-13.6}{n^2} \, eV$) (in $, V$)

When radiation is incident on a photoelectron emitter,the stopping potential is found to be $9 \ V$. If $e/m$ for the electron is $1.8 \times 10^{11} \ C \ kg^{-1}$,the maximum velocity of the ejected electrons is:

When light of intensity $1 \ W/m^2$ and wavelength $5 \times 10^{-7} \ m$ is incident on a surface, it is completely absorbed. If $100$ photons emit one electron and the surface area is $1 \ cm^2$, what is the photoelectric current?

In the case of photoelectric emission from a certain metal,the cutoff frequency is $v$. If radiation of frequency $2v$ is incident on the metal plate,the maximum possible velocity of the emitted electrons will be ($m=$ mass of electron).

$A$ photon of energy $3.4 \ eV$ is incident on a metal having work function $2 \ eV$. The maximum kinetic energy $(K.E.)$ of photo-electrons is equal to ........... $eV$. (in $eV$)