Two light beams of different frequencies,having energies $1 \ eV$ and $2.5 \ eV$ respectively,are incident on a metal surface with a work function of $0.5 \ eV$. The ratio of the maximum kinetic energies of the emitted photoelectrons is:

The light of wavelength $\lambda$ is incident on the surface of a metal with work function $\phi$ and emits electrons. What is the maximum velocity of the emitted electrons? [Given: $m=$ mass of electron,$h=$ Planck's constant,$c=$ velocity of light]

On a photosensitive material,when the frequency of incident radiation is increased by $20 \%$,the maximum kinetic energy of emitted photoelectrons increases from $0.4 \ eV$ to $0.7 \ eV$. The work function of the material is (in $eV$)

The work function of a surface of a photosensitive material is $6.2\, eV$. The wavelength of the incident radiation for which the stopping potential is $5\, V$ lies in the

The figure shows the plot of stopping potential $(V_0)$ versus $(1/\lambda)$ for three different metals. If $\phi$ is the work function,then which of the following is correct?

When a photon of energy $hv$ is incident on an aluminum plate (work function $E_0$),photoelectrons with maximum kinetic energy $K$ are emitted. What will be the maximum kinetic energy of the emitted photoelectrons when a photon of energy $2hv$ is incident on the same aluminum plate?