# 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

${E}^{\text{*"(""^1 S) ~~ "166259 cm}} ^ \left(- 1\right)$

for the $1 {s}^{1} 2 {s}^{1}$ configuration.

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$2 s$

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$1 s$

We are told that $\text{He}$ can be excited from $1 {s}^{2}$ to $1 {s}^{1} 2 {p}^{1}$ by a $\text{58.44 nm}$ excitation source.

$\underline{\textcolor{w h i t e}{\uparrow \downarrow}} \text{ "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))" "" "" "ul(color(red)(uarr) color(white)(darr))" "ul(color(white)(uarr darr))" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" ")" "" "" "underbrace(" "" "" "" "" "" "" "" }}$
$\text{ "" "" "2p" "" "" "" "" "" "" "" "" "" "" } 2 p$

$\underline{\textcolor{w h i t e}{\uparrow \downarrow}} \text{ "" "" "" "" "" "" "" "" } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$2 s \text{ "" "" "" "" "=>" "" "" "" } 2 s$

$\underline{\uparrow \textcolor{red}{\downarrow}}$$\text{ "" "" "" "" "" "" "" "" }$$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$1 s \text{ "" "" "" "" "" "" "" "" "" "" } 1 s$

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):

The state ${\text{4857.457 cm}}^{- 1}$ below is the $1 {s}^{1} 2 {s}^{1}$ configuration, or the excited state ""^1 S, with energy ${E}^{\text{*"(""^1 S) ~~ "166277.440 cm}} ^ \left(- 1\right)$.

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$2 s$

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$1 s$

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"

$= {\text{171115.674 cm}}^{- 1}$

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.