A light whose frequency is equal to `6xx10^(14)Hz` is incident on a metal whose work function is `2eV(h=6.63xx10^(-34)Js,1eV=1.6xx10^(-19)J)`. The maximum energy of electrons emitted will be:

Light of frequency 6 xx 10^(14) Hz is incident on a metal whose work function is 2eV (h=6.63 xx 10^(-34) Js,leV =1.6 xx 10^(-19)J) . The maximum energy of electrons emitted will be:

The work function of carsium metal is 2.14eV. When light of frequency 6 xx10^14 Hz is incident on the metal surface, photoemission of electrons occurs. (h=6.6 xx 10 ^-34 js) :-Calculate the maximum kinetic energy of the emitted electrons.

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

Light of frequency 4 xx 10^(14)Hz is incident on a metal surface of work function 2.14 eV, resulting in photoemission of electrons. The maximum kinetic energy of the emitted electrons is [h = 6.63 xx 10^(-34)J-s]

Frequency of photon having 100 eV energy is ……Hz [ h=6.62xx10^(-34)Js,1eV=1.6xx10^(-19)Js]

The work function of caesium is 2.14 eV. When light of frequency 6xx10^(14)Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons. (b) stopping potential and (c) maximum speed of the emitted photoelectrons. given , h=6.63xx10^(-34)Js, 1eV=1.6xx10^(-19)J, c=3xx10^(8)m//s .