Light of frequency $4\nu_0$ is incident on a metal surface with a threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is:

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

$K_1$ and $K_2$ are the maximum kinetic energies of photoelectrons emitted from a surface of a given material for light of wavelengths $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2\lambda_2$, then the work function of the material is given by:

The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

Energy required to remove an electron from an aluminum surface is $4.2 \text{ eV}$. If light of wavelength $2000 \text{ Å}$ falls on the surface,the velocity of the fastest electron ejected from the surface will be:

The work function of a photosensitive metal surface is $\phi$. When a photon of energy $3\phi$ is incident on the surface,a photoelectron with a maximum velocity of $6.6 \times 10^6 \ m/s$ is emitted. If the energy of the incident photon is increased to $9\phi$,the maximum velocity of the emitted photoelectron will be ..........