Light rays consisting of photons with energy $1.8 \ eV$ are incident on a metal surface with a work function of $1.2 \ eV$. What is the stopping potential required to stop the emitted electrons in $eV$?

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on a photosensitive surface. If only $1 \%$ of the incident photons emit photoelectrons,calculate the number of photoelectrons emitted per second from a surface area of $1 \ cm^2$.

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be:

Monochromatic radiation emitted when an electron in a hydrogen atom jumps from the first excited state to the ground state irradiates a photosensitive material. The stopping potential is measured to be $3.57 \; V$. The threshold frequency of the material is ......... $\times 10^{15} \; Hz$.

When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$,the stopping potential is $3 V_{s}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential becomes $V_{s}$. The value of the threshold wavelength for photoelectric emission is:

In the experiment of $P.E.E.$,the $KE_{max}$ of an electron is $K_0$. If the frequency is increased by a factor of $n_1$,then the $KE_{max}$ becomes $n_2K_0$. Find the work function.