The variation of stopping potential $V_{0}$ with frequency $\nu$ of incident radiation for a given photosensitive material is a straight line [frequency $\nu$ of incident radiation is greater than threshold frequency $\nu_{0}$]. The slope of this line is . . . . . . .

When a photon of wavelength $3000 \ \mathring{A}$ is incident on a metal surface,the energy of the emitted electron is $0.5 \ eV$. If a photon of wavelength $2000 \ \mathring{A}$ is incident on the same metal surface,the energy of the emitted electron will be:

$A$ beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10\%$ of the incident photons effectively produce photoelectrons,then find the current due to these electrons in $\mu A$. (Given: $hc = 1240\,eV\cdot nm$,$e = 1.6 \times 10^{-19}\,C$)

In the photoelectric effect,the threshold wavelength of sodium is $5000 \ \mathring A$. Find its work function in $eV$. $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

$(i)$ 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?

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