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?

Monochromatic light of wavelength $\lambda = 4770 \ \mathring{A}$ is incident separately on the surface of four different metals $A, B, C$ and $D$. The work functions of $A, B, C$ and $D$ are $4.2 \ \text{eV}, 3.7 \ \text{eV}, 3.2 \ \text{eV}$ and $2.3 \ \text{eV}$, respectively. From which of these metals will electrons be emitted?

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photoelectrons (of mass $m$) coming out are respectively $v_1$ and $v_2$,then:

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?

$A$ photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ (where $\lambda < \frac{hc}{\phi_0}$). The fastest photoelectron has a de Broglie wavelength $\lambda_d$. $A$ change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_d$ in $\lambda_d$. Then the ratio $\frac{\Delta \lambda_d}{\Delta \lambda}$ is proportional to:

The photoelectric threshold wavelength for a metal surface is $6600 \, \mathring{A}$. The work function for this is: