Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be:

The work functions of metal $A$ and $B$ are in the ratio $1: 2$. If light of frequency $f$ and $2f$ is incident on surface $A$ and $B$ respectively,then the ratio of the maximum kinetic energies of the emitted photoelectrons is:

In a photoelectric experiment,three different lights are incident on a metal with a work function of $1.5 \ eV$. Light $A$ has a wavelength of $200 \ nm$ and an intensity of $1.8 \ W/m^2$,light $B$ has a wavelength of $400 \ nm$ and an intensity of $1 \ W/m^2$,and light $C$ has a wavelength of $600 \ nm$ and an intensity of $0.5 \ W/m^2$. The photocurrent versus voltage is measured. Which graphs correspond to which light?

$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda/3)$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $4$ times that in the first case, the work function of the surface of the material is ($h$ = Planck's constant, $c$ = speed of light).

In the photoelectric emission process from a metal with a work function of $1.8\, eV$,the kinetic energy of the most energetic electrons is $0.5\, eV$. The corresponding stopping potential is ......... $V$.