# 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#

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

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.