Question

What is the wavelength of light that causes an electronic transition for `"Li"^(2+)` if `n_f - n_i = 2` and `n_i + n_f = 4`?

Answer 1

I got `"11.39 nm"`, for a change in energy of `"108.88 eV"`.

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Well, we first dissect the minor puzzle here... Your system of equations is:

`n_f - n_i = 2`

`n_i + n_f = 4`

where:

• `n_f` is your destination energy level in a hydrogen-like atom.
• `n_i` is the energy level from which the electron started transitioning in that hydrogen-like atom.

By simple substitution,

`n_f = 2 + n_i`

`=> n_i + 2 + n_i = 4`

`=> ul(n_i = 1)`

`=> ul(n_f = 3)`

Indeed, `3 - 1 = 2` and `1 + 3 = 4`. Anyways...

Now that we know the electronic transition is `color(green)(n = 1 -> 3)`, we invoke the Rydberg equation for hydrogen-like atoms, i.e. `"He"^(+)`, `"Li"^(2+)`, `"Be"^(3+)`, etc.:

`bb(DeltaE = -Z^2 cdot "13.61 eV"(1/n_f^2 - 1/n_i^2))`

where:

• `DeltaE` is the change in energy due to the transition.
• `-"13.61 eV"` is the ground-state energy of hydrogen atom. The magnitude, `R_H ~~ "13.61 eV"`, is also called the Rydberg constant.
• `Z` is the atomic number.

The wavelength `lambda` is gotten by converting the resultant change in energy, which corresponds to the energy of the emitted photon during the electronic relaxation. In other words:

`bb(DeltaE = E_"photon" = hnu = (hc)/(lambda))`

where:

• `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant.
• `c = 2.998 xx 10^(8) "m/s"` is the speed of light.
• `nu` is the frequency of the photon in `"s"^(-1)`.

And so, the Rydberg equation for wavelength is:

`bb((DeltaE)/(hc) = 1/lambda = -(Z^2 cdot "13.61 eV")/(hc)(1/n_f^2 - 1/n_i^2))`

Knowing that, the wavelength will be:

`lambda = [-(3^2 cdot 13.61 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV"))/((6.626 xx 10^(-34) "J"cdot"s" xx 2.998 xx 10^(8) "m/s"))(1/3^2 - 1/1^2)]^(-1)`

`=` `1.139 xx 10^(-8) "m"`

In nicer units, we then have:

`color(blue)(lambda) = 1.139 xx 10^(-8) cancel"m" xx (10^9 "nm")/(cancel"1 m")`

`=` `color(blue)(ul"11.39 nm")`

Just to be sure you get the right value, the energy you get, which is easier to calculate, would have been:

`color(blue)(DeltaE) = -Z^2 cdot "13.61 eV"(1/3^2 - 1/1^2)`

`= -3^2 cdot "13.61 eV"(1/9 - 1/1)`

`=` `color(blue)(ul"108.88 eV")`

And to check, we look on NIST. The energy levels for `n = 1->3` in `"Li"^(2+)` are listed as:

These are in `"cm"^(-1)`. Take the first one for example (ignore the slight differences in the two choices).

`lambda_(ref) = 1/("877919.12077 "cancel("cm"^(-1))) xx (10^7 "nm")/(cancel"1 cm")`

`=` `"11.3906 nm"` `~~` `"11.39 nm"`

So, the wavelength we got is correct.