The threshold frequency of potassium is $3 \times 10^{14} \ Hz$. The work function is ...... .

In a photoelectric experiment,a monochromatic light is incident on the emitter plate $E$,as shown in the figure. When switch $S_1$ is closed and switch $S_2$ is open,the photoelectrons strike the collector plate $C$ with a maximum kinetic energy of $1 eV$. If switch $S_1$ is open and switch $S_2$ is closed and the frequency of the incident light is doubled,the photoelectrons strike the collector plate with a maximum kinetic energy of $20 eV$. The threshold wavelength of the emitter plate is (in $Å$)

The metallic surface is illuminated with monochromatic light of wavelength $\lambda$ and the stopping potential for the photoelectric current is $5V_0$. When the same metallic surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. What is the threshold wavelength for the surface?

Two light waves of wavelengths $600 \,nm$ and $200 \,nm$ are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is $\frac{1}{3}$ of the maximum velocity of the photoelectrons produced due to the other wavelength. The work function of the metal is:

Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$,the wavelength must be changed to $\lambda'$ where

$(i)$ In the explanation of the photoelectric effect,we assume one photon of frequency $f$ collides with an electron and transfers its energy. This leads to the equation for the maximum kinetic energy $E_{max}$ of the emitted electron as $E_{max} = hf - \phi_0$ (where $\phi_0$ is the work function of the metal). If an electron absorbs $2$ photons (each of frequency $f$),what will be the maximum energy for the emitted electron?
$(ii)$ Why is this fact (two-photon absorption) not taken into consideration in our discussion of the stopping potential?