In an experiment on the photoelectric effect,the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \; V \cdot s$. Calculate the value of Planck's constant.

In the graph given below,if the slope is $4.12 \times 10^{-15} \, V-s$,then the value of $'h'$ should be:

$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14 \text{ eV}$ and the stopping potential is $2 \text{ V}$, what is the wavelength of the electromagnetic wave (in $\text{ nm}$)? (Given $hc = 1242 \text{ eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)

The work function of a material is $4.0 \ eV$. The maximum wavelength of light that can cause the emission of photoelectrons from the material is ............ $nm$.

When light of wavelength $300 \ nm$ is incident on a photoelectric emitter,photoelectrons are emitted. For another emitter,light of wavelength $600 \ nm$ is sufficient for photoemission. What is the ratio of the work functions of the two emitters?