The threshold frequency for a certain metal is $3.3 \times 10^{14} \;Hz$. If light of frequency $8.2 \times 10^{14} \;Hz$ is incident on the metal,predict the cutoff voltage (in $eV$) for the photoelectric emission.

When light of wavelength $\lambda$ incidents on a photosensitive material,photoelectrons are emitted. If the wavelength of the incident light is reduced by $50 \%$,the maximum kinetic energy of the emitted photoelectrons becomes $3$ times the initial maximum kinetic energy. The work function of the material is ($h$ - Planck's constant,$c$ - Speed of light in vacuum).

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy,for the same wavelength of incident radiation?

By the photoelectric effect,Einstein proved:

If light of wavelength $6600 \ \mathring A$ is incident on a metal surface with a work function of $2 \ eV$,what will be the maximum kinetic energy of the emitted photoelectrons?

In a photoelectric experiment,the stopping potential was measured to be $V_1$ and $V_2$ volts with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The value of $V_2$ is: [where $\phi=$ work function,$e=$ electronic charge]