When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with kinetic energy of `1.68 xx 10^(5) J "ml"^(-1)`. What is the minimum energy needed to remove an electron from sodium ? What is the maximum wavelength that will cause a photoelectron to be emitted.

Explanation

Energy of the striking photon,
`E = hv = (hc)/(lambda)`
`h - 6.626 xx 10^(-34) J s, c = 3.0 xx 10^8 m s^(-1)`
`lambda = 300 nm = 300 xx 10^(-9) m`
`:. E = (6.626 xx 10^(-34) J s xx 3.0 xx 10^8 ms^(-1))/(300 xx 10^(-9) m)`
`= 6.626 xx 10^(-19) J`
Kinetic energy of emitted elelctrons = `1.68 xx 10^6 J mol^(-1)`
Kinetic energy of emitted one electron = `(1.68 xx 10^(5))/(6.022 xx 10^(23))`
`= 2.79 xx 10^(-19) J`
Now,
Energy of striking photon = minimum energy required to eject electron + Kinetic energy of electrons or Minimum energy required for ejection of an electron
`= 6.626 xx 10^(-19) J - 2.79 xx 10^(-19) J`
`= 3.84 xx 10^(-19) J`
The wavelength which will cause photoelectron emission,
`lambda = (hc)/E`
`=(6.626 xx 10^(-34) J s xx 3.0 xx 10^8 ms^(-1))/(3.84 xx 10^(-19) J)`
`= 5.17 xx 10^(-7) m`
or `= 517 xx 10^(-9) m = 517 nm`.