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 $eV$?

If the wavelengths of incident radiation are $2500 \ \mathring A$ and $5000 \ \mathring A$ respectively,and the work function of the metal surface is $2 \ eV$,find the approximate ratio of the stopping potentials for the emitted photoelectrons. (in $:1$)

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength $\lambda_1$ and $K_2$ corresponding to $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

When a photon with energy $8\ eV$ is incident on a metal surface having a threshold frequency of $1.6 \times 10^{15}\ Hz$,the maximum kinetic energy of the emitted photoelectrons is ............ $eV$. (Given: $h = 6.6 \times 10^{-34}\ J\cdot s$,$1\ eV = 1.6 \times 10^{-19}\ J$)

When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is

When a metal is irradiated by monochromatic light,the maximum kinetic energy of the photoelectrons is $1.2 \ eV.$ If the frequency of the light is increased by $50 \%$,then the maximum kinetic energy of the photoelectrons is $3.6 \ eV.$ Evaluate the work function of the metal.