A photon of energy `8 eV` is incident on a metal surface of threshold frequency `1.6 xx 10^(15) Hz`, then the maximum kinetic energy of photoelectrons emitted is `( h = 6.6 xx 10^(-34) Js)`

A photon of energy 8 eV is incident on metal surface of threshold frequency 1.6 xx 10^(15) Hz , The maximum kinetic energy of the photoelectrons emitted ( in eV ) (Take h = 6 xx 10^(-34) Js ).

A photon of energy 8 eV is incident on a metal surface of threshold frequency 1.6xx10^(15)Hz . What is the K.E. of the ejected photoelectrons in eV ? [h=6xx10^(-34)Js]

A photon of frequency v is incident on a metal surface whose threshold frequency is v_(0) . The kinetic energy of the emitted photoelectrons will be

Light of frequency 4 v_(0) is incident on the metal of the threshold frequency v_(0) . The maximum kinetic energy of the emitted photoelectrons is

Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV . The minimum kinetic energy of the emitted photo - electrons will be

If a photoemissive surface has a threshold frequency of . 6 xx 10^(14)Hz , calculate the energy of the photons in eV. Given h=6.6 xx 10^(-34)Js .

If a photoemissive surface has a threshold frequency of 3 xx 10^(14) Hz, calculate the energy of the photons in eV.

A photon of energy 2.16eV is incident on the surface of a metal of photoelectric work function 1.25eV. If the maximum velocity of a photoelectron emitted is 0.566 xx 10^(6) ms^(-1) , calculate the mass of the electron.

If the work function for a certain metal is 3.2 xx 10^(-19) joule and it is illuminated with light of frequency 8 xx 10^(14) Hz . The maximum kinetic energy of the photo-electrons would be (h = 6.63 xx 10^(-34) Js)

Light of frequency 8 xx 10^(15) Hz is incident on a substance of photoelectric work function 6.125 eV . The maximum kinetic energy of the emitted photoelectrons is