Light from the Paschen series of a hydrogen atom is able to eject photoelectrons from a metal. Then the work function of the metal is:

In a photoelectric effect experiment,if the frequency of light is doubled,the stopping potential will

Light of wavelength $4000 \,\mathring A$ falls on a photosensitive metal and a negative $2 \,V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 \times 10^{-34} \,Js, \,e = 1.6 \times 10^{-19} \,C, \,c = 3 \times 10^8 \,m/s)$.

In an experiment on the photoelectric effect,the wavelength of the incident radiation is $\lambda$. The wavelength of the incident radiation is reduced to $\frac{1}{3}$ of the initial value,and the maximum kinetic energy of the photoelectron is observed to be $n$ times the previous value. The threshold wavelength for the metal plate is:

$(i)$ 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?

$A$ 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 $\text{eV}$)