Answer to Question #207605 in Quantum Mechanics for shubham

Question

Light of frequency 6 × 1014 Hz incident on a metal surface ejects 

photoelectrons with a maximum kinetic energy of 2 × 10–19 J. Calculate 

the work function of the metal, giving your answer in eV.

Answer 1

The incident (photon) energy provided as (Planck's Constant times frequency) minus the energy that "binds" the electron to the metal will be equal to the maximal kinetic energy of, say, one electron released (the work function)

So, to summarize:

$K= hf-\phi$

$2 × 10^{–19} J = 6.63*10^{-34}* 6 × 10^{14} Hz-\phi$

$2 × 10^{–19} = 6.63* 6 × 10^{14-34}-\phi$

$2 × 10^{–19} = 39.78 × 10^{-20}-\phi$

$2 × 10^{–19} - 39.78 × 10^{-20}=-\phi$

$2 × 10^{–19} - 3.978 × 10^{-19}=-\phi$

$10^{–19}(2 - 3.978 )=-\phi$

$\phi=1.978 *10^{–19} eV$