Light of incident frequency $3$ times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{4}\right)^{\text{th}}$ and intensity is tripled,then the photoelectric current will

$A$ photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$,then the radius $r$ is given by,(in the usual notation)

Photoelectrons are emitted from a photosensitive surface for light of wavelengths $\lambda_{1} = 360 \ nm$ and $\lambda_{2} = 600 \ nm$. What is the ratio of the work functions for the lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$?

$A$ 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 $\text{eV}$)

The difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work functions $\phi_A = 9 \, eV$ and $\phi_B = 4.5 \, eV$ in $nm$ is: (Given,$hc = 1242 \, eV \, nm$)

When a metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is: