Photons of energy $2.4 \text{ eV}$ and wavelength $\lambda$ fall on a metal plate and release photoelectrons with a maximum velocity $v$. By decreasing $\lambda$ by $50 \%$, the maximum velocity of photoelectrons becomes $3 v$. The work function of the material of the metal plate is (in $\text{ eV}$)

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $3V_0$. When 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:

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 work function of metals is in the range of $2 eV$ to $5 eV$. Find which of the following wavelengths of light cannot be used for the photoelectric effect (in $nm$)? (Consider, Planck constant $= 4 \times 10^{-15} eVs$, velocity of light $= 3 \times 10^{8} m/s$)

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is: