What potential difference must be applied to stop the fastest photoelectrons emitted by a nickel surface under the action of ultraviolet light of wavelength `2000Å`? The work function of nickel is 5.01 eV.

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface , having work function 5.01 eV , when ultraviolet light of 200 nm falls on it , must be

No photoelectrons are emitted from a metal if the wavelength of light exceeds 6000 Å .The work function of the metal is approximately equal to:

Five volt of stopping potential is needed for the photoelectrons emitted out of a surface of work function 2.2 eV by the radiation of wavelength

Calculate the kinetic energy of a photoelectron (in eV) emitted on shining light of wavelength 6.2 xx 10^(-6)m on a metal surface. The work function of the metal is 0.1 eV ?

(a) The work function for the surface of aluminium is 4.2eV. Find potential difference will be required to stop the emission of maximum kinetic energy photoelectrons emitted by light of 1800Å wavelenght? (b) Determine the wavelength of that incident light for which stopping potential will be zero? (h=6.6xx10^(-34))

0.24eV of energy is required to stop photoelectron emitted by radiation of wavelength 2480 Å . Find work function of metal used?

The work functin of aluminium is 4.2 eV. Calcualte the Kinetic energy of the fastest & the slowest photoelectrons, the stopping potential & the cut off wavelength when light of wavelength 2000Å falls on a clean Aluminium surface.

(a) The work function for the surface of aluminium is 4.2eV. How much potential difference will be required to stop the emission of maximum energy electrons emitted by light of 2000Å wavelenght? (b) What will be the wavelength of that incident light for which stopping potential will be zero ? h=6.6xx10^(-34) Js, c=3xx10^8ms^-1 .