Short Answer Type
Long Answer TypeExplain briefly the photoelectric effect.
The phenomenon of ejection of electrons from the surface of a metal when the light of suitable frequency strikes it is called photoelectric effect and the emitted electrons are called photoelectrons.
For each metal, there is a characteristic minimum frequency called threshold frequency below which the photoelectric effect does not occur. For the photoelectric effect to occur, the striking photon should have frequency more than that of the threshold frequency. If a photon of frequency v strikes a metal atom whose threshold energy is v0, then photoelectrons will be emitted only if v > v0. Since the striking photon has energy equal to hv and minimum energy required to eject electron is hv0 (called wave function W0), then the excess of energy i.e. hv - hv0 or h(v - v0) will be imparted to the ejected electron as kinetic energy.
Hence,
K.E. of ejected electron 
where me is the mass of electron and v be its velocity.
A photon of wavelength 4 x 10-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate: (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 10-19J).
Short Answer TypeLong Answer TypeThe work function for caesium atom is 1.9 eV. Calculate: (a) the threshold wavelength and (b) the threshold frequency of radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and velocity of the ejected photoelectron.
Short Answer Type
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