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
E^"*"(""^1 S) ~~ "166259 cm"^(-1)
for the
ul(uarr color(white)(darr))
2s
ul(uarr color(white)(darr))
1s
We are told that
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
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The state
ul(uarr color(white)(darr))
2s
ul(uarr color(white)(darr))
1s
That is the state circled in red in the
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.