When radiation of wavelength $\lambda$ is incident on a photocell,the maximum velocity of the photoelectrons is $v$. What will be the maximum velocity of the photoelectrons when radiation of wavelength $3\lambda/4$ is incident on the photocell?

$A$ 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).

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

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).

Light of wavelength $3500 \ \mathring A$ is incident on two metals $A$ and $B$. If their work functions are $4.2 \ eV$ and $1.9 \ eV$ respectively,which metal will emit photoelectrons?

$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be: