In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

The work function of a photosensitive metal surface is $hv_0$. If a photon of energy $2hv_0$ is incident on this surface,electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. What will be the maximum velocity of the photoelectrons when the energy of the incident photon is increased to $5hv_0$?

$A$ beam of photons with an energy of $10.5 \ eV$ strikes a metal plate. The photoelectrons are emitted with a maximum velocity of $1.6 \times 10^6 \ m \ s^{-1}$. The work function of the metal is (Assume mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$). (in $eV$)

The threshold wavelength of tungsten is $2300 \, \mathring{A}$. If ultraviolet light of wavelength $1800 \, \mathring{A}$ is incident on it,then the maximum kinetic energy of photoelectrons would be about ............ $eV$.

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is: