Two identical metal plates show the photoelectric effect. Light of wavelength ${\lambda _A}$ falls on plate $A$ and light of wavelength ${\lambda _B}$ falls on plate $B$,where ${\lambda _A} = 2{\lambda _B}$. The maximum kinetic energy is:

According to the photoelectric effect,the plot of kinetic energy of the emitted photo-electrons from a metal versus the frequency of the incident radiation gives a straight line whose slope

When photons of energy $hf$ fall on an aluminium plate (of work function $E_0$),photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be:

$A$ photoelectric cell is connected to a source of variable potential difference and the resulting photoelectric current $(\mu A)$ is plotted against the applied potential difference $(V)$. The graph with the broken line represents one situation for a given frequency and intensity of the incident radiation. If the frequency is increased and intensity is reduced,which of the following graphs of unbroken line represents the new situation?

The work function of a metal is $2.51 \ eV$. What is its threshold frequency?

The maximum wavelength of radiation that can produce the photoelectric effect in a certain metal is $200 \ nm$. The maximum kinetic energy acquired by an electron due to radiation of wavelength $100 \ nm$ will be .............. $eV$.