He atom can be excited to #1s^1 2p^1# by #lambda=58.44 nm#. If lowest excited state for He lies #4857 cm^(-1)# below the above. Calculate the energy for lower excitation state?

1 Answer
Jul 21, 2017

#E^"*"(""^1 S) ~~ "166259 cm"^(-1)#

for the #1s^1 2s^1# configuration.

#ul(uarr color(white)(darr))#
#2s#

#ul(uarr color(white)(darr))#
#1s#


We are told that #"He"# can be excited from #1s^2# to #1s^1 2p^1# by a #"58.44 nm"# excitation source.

#ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "" "" "ul(color(red)(uarr) color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))#
#underbrace(" "" "" "" "" "" "" "" ")" "" "" "underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "2p" "" "" "" "" "" "" "" "" "" "" "2p#

#ul(color(white)(uarr darr))" "" "" "" "" "" "" "" "" "ul(color(white)(uarr darr))#
#2s" "" "" "" "" "=>" "" "" "" "2s#

#ul(uarr color(red)(darr))##" "" "" "" "" "" "" "" "" "##ul(uarr color(white)(darr))#
#1s" "" "" "" "" "" "" "" "" "" "" "1s#

https://www.researchgate.net/

By inspection of the above energy level diagram, indeed it can. That is a diagonal excitation as seen above (legal by the selection rules), going from the ground-state #""^1 S# to the #""^1 P# state, i.e. from #E(""^1 S) = "0.000 cm"^(-1)# to #E(""^1 P) = "171134.897 cm"^(-1)#:

https://physics.nist.gov/

The state #"4857.457 cm"^(-1)# below is the #1s^1 2s^1# configuration, or the excited state #""^1 S#, with energy #E^"*"(""^1 S) ~~ "166277.440 cm"^(-1)#.

#ul(uarr color(white)(darr))#
#2s#

#ul(uarr color(white)(darr))#
#1s#

That is the state circled in red in the #""^1 S# column in the energy level diagram above.

So, we know what we should get. Let's get it the normal way now.

#E(""^1 P) = 1/(58.44 cancel"nm") xx (10^9 cancel"nm")/(cancel"1 m") xx cancel"1 m"/"100 cm"#

#= "171115.674 cm"^(-1)#

And we know that the lower-excited state that the electron relaxes to has energy

#E(""^1 P) - "4857 cm"^(-1)#

So, the energy of the lower excited state with your level of precision is:

#= ul(color(blue)("166259 cm"^(-1) ~~ E^"*"(""^1 S)))#,

keeping zero decimal places.