Photoelectric effect can be explained by

In a photoemission experiment,the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_P, E_Q$ and $E_R$,respectively,and they are related by $E_P = 2E_Q = 2E_R$. In this experiment,the same source of monochromatic light is used for metals $P$ and $Q$,while a different source of monochromatic light is used for metal $R$. The work functions for metals $P, Q$ and $R$ are $4.0 \ eV$,$4.5 \ eV$ and $5.5 \ eV$,respectively. The energy of the incident photon used for metal $R$,in $eV$,is:

Photons of wavelength $\lambda$ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius $R$ by a perpendicular magnetic field having a magnitude $B$. The work function of the metal is (where,symbols have their usual meanings)

When the wavelength of radiation falling on a metal is changed from $500 \, nm$ to $200 \, nm$,the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to $..... \, eV$.

When photons of energy $h \nu$ fall on a photosensitive surface of work function $E_0$,photoelectrons of maximum kinetic energy $k$ are emitted. If the frequency of radiation is doubled,the maximum kinetic energy will be equal to ($h=$ Planck's constant).

Photoelectrons are emitted from a metal surface for frequencies $\nu_1$ and $\nu_2$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1 : k$,then the threshold frequency of the metal surface is: