When photons of energy $1 \ eV$ and $2.5 \ eV$ are incident on a metal surface with a work function of $0.5 \ eV$,what is the ratio of the maximum kinetic energies of the emitted photoelectrons?

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $3\lambda$,the stopping potential is $\frac{V}{6}$. The threshold wavelength for the surface is:

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. $A$ light source is turned on and a saturation photocurrent is recorded. An electric field is then switched on which has a vertically downward direction.

If the electron in a hydrogen atom jumps from the third Bohr orbit to the ground state directly and the difference between the energies of the two states is radiated in the form of photons. If the work function of the material is $4.1 \text{ eV}$, then the stopping potential is nearly:
$\left[\text{Energy of electron in } n^{\text{th}} \text{ orbit} = \frac{-13.6}{n^2} \text{ eV}\right]$ (in $\text{ V}$)

The work function of a substance is $4.0 \,eV$. The longest wavelength of light that can cause photoelectron emission from this substance is approximately ......... $nm$.