Question

if a photon of energy 7.08 into 10 ^ - 19 j falls on Lithium metal having work function 2.42 EV then the kinetic energy of the ejected electron will be​

Answer 1

energy of incident photons, E = 7.08 × 10^-19 J

we know, 1eV = 1.6 × 10^-19 J

so, energy of incident photon, E = (7.08 × 10^-19)/(1.6 × 10^-19) eV

= 4.425 eV

given,

work function, Φ = 2.42 eV

we know, from Einstein's photoelectric effect theory,

Kinetic energy = energy of incident photon - work function

= E - Φ

= 4.425 - 2.42

= 2.005 eV

hence, kinetic energy of ejected electron will be 2.005 eV