Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

$A$ photon of wavelength $3315 \ \text{Å}$ falls on a photocathode and an electron of energy $3 \times 10^{-19} \ \text{J}$ is ejected. The threshold wavelength of the photon is [Planck's constant $(h)$ $= 6.63 \times 10^{-34} \ \text{J-s}$, velocity of light $(c)$ $= 3 \times 10^{8} \ \text{m/s}$]. (in $\text{Å}$)

When ultraviolet radiation of a certain frequency falls on a potassium target,the photoelectrons released can be stopped completely by a retarding potential of $0.6 \, V$. If the frequency of the radiation is increased by $10 \%$,this stopping potential rises to $0.9 \, V$. The work function of potassium is ........ $eV$.

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?

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