Calculate the vel,ocity of electron ejected from a pltinum when radiation of `200` nm falls on it .The work function of platinum in `5 eV,(1 eV = 1.6 xx 10^(-19)J)`

Calculate the velocity of electron ejected from a platium surface when radiation of 200 nm falls on it .The work function of platinum is 5 eV(1 eV = 1.6 xx 10^(-19) J)

Calculate the velocity of electron ejected from a platium surface when radiation of 200 nm falls on it .The work function of platinum is 5 eV(1 eV = 1.6 xx 10^(-19) J)

Calculate the velocity of electron ejected from platinum surface when radiation of 200 nm falls on it. Work function of platinum is 5eV. (1eV=1.6xx10^-19J)

Energy required to stop the ejection of electrons from certain metal plate was found to be 0.2 eV when an electromagnetic radiation of 330 nm was made to fall on it the work function of the metal is

Energy required to stop the ejection of electrons from certain metal plate was found to be 0.2 eV when an electromagnetic radiation of 330 nm was made to fall on it the work function of the metal is

Calculate the longest wavelength of incident light on platinum for which photo electrons are just emitted from platinum (work function for platinum = 6.2 eV).

Find the maximum velocity of photoelectrons emitted by radiation of frequency 3 xx 10^(15) Hz from a photoelectric surface having a work function of 4.0 eV. Given h=6.6 xx 10^(-34) Js and 1 eV = 1.6 xx 10^(-19) J.

Find the maximum velocity of photoelectrons emitted by radiation of frequency 3 xx 10^(15) Hz from a photoelectric surface having a work function of 4.0 eV. Given h=6.6 xx 10^(-34) Js and 1 eV = 1.6 xx 10^(-19) J.

Calculate the wavelength and energy of radiation for the elctronic transition form infinity to ground state for one H-atom . Given e_1 =- 13. 6 eV ( 1 1 eV = 1,. 6 xx 10^(-19) J) .

A stream of photons of energy 10.6 eV and intensity 2.0 W m^(-2) is incident on a platinum surface. Area of the surface is 1.0 xx 10^(-4) m^(2) and its work function is 5.6 eV. 0.53 % of incident photons emit photoelectrons. Find the number of photoelectrons emitted per second and maximum and minimum energies of the emitted photoelectrons in eV. (1eV = 1.6 xx 10^(-19)J)