If the ionization potential for hydrogen atom is 13.6eV, then the wavelength of light required for the ionization if hydrogen atom would be ?

1 Answer
Truong-Son N.
Jun 5, 2018

#"91.2 nm"#.

Well, photons are incoming particles for ionizing atoms, and they have energy

#E = hnu = (hc)/lambda#

where:

  • #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant.
  • #c = 2.998 xx 10^8 "m/s"# is the speed of light.
  • #lambda# is the wavelength in #"m"#.
  • #nu# is the frequency in #"s"^(-1)#.

The ionization energy is just the energy needed to excite the electron out of the ground state (#n=1#) into the continuum (#n -> oo#), i.e. #Deltan = 1 -> oo#.

Thus, since there are about #1.602 xx 10^(-19) "J/eV"#, the wavelength is

#color(blue)(lambda) = (hc)/(DeltaE_(1->oo))#

#= (6.626 xx 10^(-34) cancel"J"cdotcancel"s" cdot 2.998 xx 10^8 "m/"cancel"s")/(13.6 cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/cancel("1 eV"))#

#= 9.12 xx 10^(-8) "m" = color(blue)("91.2 nm")#

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