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

When a light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of emitted photoelectrons is $V$. If the incident wavelength is changed to $2\lambda / 3$,the maximum speed of the emitted photoelectrons will be:

Light of wavelength $5000 \ \mathring A$ and intensity $4.68 \ mW/cm^2$ is incident on a photosensitive surface. If only $5\%$ of the incident photons produce photoelectrons,what is the number of photoelectrons emitted per unit area per unit time?

When the wavelength of incident radiation on a metal surface is reduced from $\lambda_1$ to $\lambda_2$,the kinetic energy of the emitted photoelectrons is tripled. Find the work function of the metal. [$h =$ Planck's constant,$c =$ velocity of light]

Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$,respectively,are incident on a metallic surface. If $\lambda_{1} = 3 \lambda_{2}$,then:

Assertion : When ultraviolet light is incident on a photocell,its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X-$ rays,both $V_0$ and $K_{max}$ increase.
Reason : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.