When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

The work function of a photosensitive metal surface is $1.1 \ eV$. Two light beams of energies $1.5 \ eV$ and $2 \ eV$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

How does the maximum kinetic energy of a photoelectron depend on the frequency of incident radiation? On which factors does it not depend?

When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for photoelectric current is $V_0$. When the same surface is illuminated by light of frequency $\frac{v}{2}$,the stopping potential is $\frac{V_0}{4}$,the threshold frequency of photoelectric emission is

For two different photosensitive materials having work functions $\phi$ and $2 \phi$ respectively,which are illuminated with light of sufficient energy to emit electrons,if the graph of stopping potential $(V_s)$ versus frequency $(\nu)$ is drawn for these two materials,what is the ratio of the slopes of the graphs for these two materials?

If the maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]