Question

Determining energy difference (in kJ/mol) between excited and ground states of Li+?

Excited lithium ions emit radiation at a wavelength of 670.8 nm in the visible region of the spectrum. (This gives the characteristic red colour in the flame test for this element).

a) What is the frequency of this radiation?

(What I have so far):

`v = c/lambda = (2.998times10^8 ms^-1) / (6.708times10^-7 m)`

`v = 4.469times10^14 s^-1`

b) If this red light is caused by an electronic transition from an excited state to the ground state of Li+, calculate the energy difference, in kJ/mol, between these two states in Li+.

Part b is primarily what I cannot figure out.

Answer 1

Well, the question doesn't make physical sense, because the difference between these two states (`1s^1 2s^1` of `"Li"^(+)` to `1s^2` of `"Li"^(+)`) is about `"476034.98 cm"^-1`, or the equivalent of `"59.0 eV"`.

https://physics.nist.gov/PhysRefData/ASD/levels_form.html
[Search "Li II" without the quotes.]

The frequency (which gives an energy difference magnitude of `"1.85 eV"`) couldn't possibly correspond to the relaxation from `1s^1 2s^1` to `1s^2` of `"Li"^+`.

What this should have said is:

If this red light is caused by an electronic transition from an excited state to the ground state of `color(blue)(Li)`, calculate the energy difference, in `kJ//mol`, between these two states in `color(blue)(Li)`.

Physically, we observe the emission from the `color(blue)(1s^2 2p^1)` excited state to the `color(blue)(1s^2 2s^1)` ground state of `bb"Li"` neutral, which is precisely `"14903.660 cm"^-1` (corresponding to `"1.85 eV"`) as it should be!

`underbrace(ul(cancel(uarr) color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr)))_(2p)`

`color(white)(;.)color(red)|`
`color(white)(..)color(red)darr`

`underbrace(ul(uarr color(white)(darr)))_(2s)`

https://physics.nist.gov/PhysRefData/ASD/levels_form.html
[Search "Li I" without the quotes.]

---

Having cleared that up, let's first go from the frequency to the energy. Here we just use the fact that a quantum of energy is `hnu`, where `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant.

`DeltaE = E_"photon" = hnu`

`= 6.626 xx 10^(-34) "J" cdot "s" xx 4.469 xx 10^14 "s"^(-1)`

`= 2.96 xx 10^(-19) "J"`

In terms of `"eV"`, we would have:

`color(blue)(DeltaE_(""^(3)S -> ""^(1) S)) = -2.96 xx 10^(-19) cancel"J" xx ("1 eV")/(1.602 xx 10^(-19) cancel"J") = color(blue)(-"1.85 eV")`

In terms of `"kJ/mol"`, we would have:

`color(blue)(DeltaE_(""^(3)S -> ""^(1) S)) = -2.96 xx 10^(-19) cancel"J" xx ("1 kJ")/(1000 cancel"J") xx (6.0221413 xx 10^23)/("mol")`

`=` `color(blue)(-"178.33 kJ/mol")`