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.

The work function of a metal is 4.2 eV If radiation of 2000 Å fall on the metal then the kinetic energy of the fastest photoelectrn is

Light of wavelength 200 nm shines on an aluminum surface, 4.20 eV is required to eject an electron. What is the kinetic energy of (a) the fastest and (b) the slowest ejected electrons? (c) What is the stopping potential for this situation? (d) What is the cutoff wavelength for aluminum?

Light of wavelength 2000 Å falls on an aluminium surface . In aluminium 4.2 e V of energy is required to remove an electron from its surface. What is the kinetic energy , in electron volt of (a) the fastest and (b) the slowest emitted photo-electron . ( c) What is the stopping potential ? (d) What is the cut - off wavelength for aluminum? (Plank's constant h = 6.6 xx 10^(-34) J-s and speed of light c = 3 xx 10^(8) m s^(-1).

The work function for aluminium is 4.125 eV. The cut-off wavelength for photoelectric effect for aluminium will be

The photoelectric work function of a metal is 3 eV . The maximum kinetic energy of the emitted electrons when light of wavelength 3000 Å falls on it is