The threshold frequency of a metal with work function $6.63 \ eV$ is:

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $4 V_0$. When the same surface is illuminated with light of wavelength $3 \lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is

The work function of a photosensitive material is $4.0 \ eV$. The longest wavelength of light that can cause photon emission from the substance is approximately $...... \ nm$.

$(i)$ 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?

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photo-electrons would be (given $h = 6.63 \times 10^{-34} \ J \cdot s$):

If the frequency of light falling on a photosensitive material doubles, which of the following is true?