When radiation is incident on a photoelectron emitter,the stopping potential is found to be $9 \ V$. If $e/m$ for the electron is $1.8 \times 10^{11} \ C \ kg^{-1}$,the maximum velocity of the ejected electrons 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?

When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W$), then the photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2}$, the stopping potential changes to $V_{2}$. Therefore, the charge of an electron $e$ is given by:

When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$,the stopping potential is $3 V_{s}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential becomes $V_{s}$. The value of the threshold wavelength for photoelectric emission is:

When the wavelength of an incident photon is decreased, then:

The work function for a certain metal is $4.2 \; eV$. Will this metal give photoelectric emission for incident radiation of wavelength $330 \; nm$?