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 1.6 xx 10^(15) Hz , then the maximum kinetic energy of photoelectrons emitted is ( h = 6.6 xx 10^(-34) Js)

A photon of energy 10 eV is incident on a metal surface of threshold frequency 1.6 xx 10^15 Hz . The kinetic energy of the photoelectron (in eV) is (assume h = 6 x 10^-34 )

A photon of energy 8 eV is incident on a metal surface of threshold frequency 1.6 xx 10^15 Hz . The K.E. of the photoelectrons emitted (in eV) is (Take h = 6 xx 10^-34 J - s )

A photon of energy 8 eV is incident on a metal surface of threshold frequency 1.6 xx 10^15 Hz . The K.E. of the photoelectrons emitted is (in eV). Take h as 6 xx 10^-34 J s .

A photon of energy 8eV is incident on a metal plate with threshold frequency 1.6 xx 10^(15) Hz. The maximum kinetic energy of the emitted photo electrons is (given 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

A photon having 8 eV energy is incident on metal suraface with threshold frequency of 1.6xx10^(15) Hz .Maximum kinetic energy of photoelectron emitted will be……. h=6.6xx10^(-34)JS,c=3xx10^(8)m//s eV=1.6xx10^(-19)JS