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

The work function of $Zn$ is $4.25 \ eV$. The threshold frequency corresponds to which region of the electromagnetic spectrum?

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $3V_0$. When the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

$A$ beam of light of wavelength $\lambda$ falls on a metal having work function $\phi$ placed in a magnetic field $B$. The most energetic electrons,moving perpendicular to the field,are bent in circular arcs of radius $R$. If the experiment is performed for different values of $\lambda$,then the $B^2$ vs. $\frac{1}{\lambda}$ graph will look like (keeping all other quantities constant):

What is the threshold wavelength in $nm$ for a metal with a work function of $4.0 \ eV$?

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other,the maximum velocities of the photoelectrons emitted in the two cases are $v_1$ and $v_2$ respectively. The ratio $v_1: v_2$ is