In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

Sodium and copper have work functions $2.3 \ eV$ and $4.5 \ eV$ respectively. Then the ratio of their threshold wavelengths is nearest to

$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

The work function of a photosensitive metal surface is $hv_0$. If a photon of energy $2hv_0$ is incident on this surface,electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. What will be the maximum velocity of the photoelectrons when the energy of the incident photon is increased to $5hv_0$?

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $k : 1$, then what is the threshold frequency of the metallic surface?

The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.