# 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?

Aug 8, 2017

I got $\text{11.39 nm}$, for a change in energy of $\text{108.88 eV}$.

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,

${n}_{f} = 2 + {n}_{i}$

$\implies {n}_{i} + 2 + {n}_{i} = 4$

$\implies \underline{{n}_{i} = 1}$

$\implies \underline{{n}_{f} = 3}$

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

Now that we know the electronic transition is $\textcolor{g r e e n}{n = 1 \to 3}$, we invoke the Rydberg equation for hydrogen-like atoms, i.e. ${\text{He}}^{+}$, ${\text{Li}}^{2 +}$, ${\text{Be}}^{3 +}$, etc.:

$\boldsymbol{\Delta E = - {Z}^{2} \cdot \text{13.61 eV} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)}$

where:

• $\Delta E$ is the change in energy due to the transition.
• $- \text{13.61 eV}$ is the ground-state energy of hydrogen atom. The magnitude, ${R}_{H} \approx \text{13.61 eV}$, 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:

$\boldsymbol{\Delta E = {E}_{\text{photon}} = h \nu = \frac{h c}{\lambda}}$

where:

• $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant.
• $c = 2.998 \times {10}^{8} \text{m/s}$ is the speed of light.
• $\nu$ is the frequency of the photon in ${\text{s}}^{- 1}$.

And so, the Rydberg equation for wavelength is:

$\boldsymbol{\frac{\Delta E}{h c} = \frac{1}{\lambda} = - \frac{{Z}^{2} \cdot \text{13.61 eV}}{h c} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)}$

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)

$=$ $1.139 \times {10}^{- 8} \text{m}$

In nicer units, we then have:

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

$=$ $\textcolor{b l u e}{\underline{\text{11.39 nm}}}$

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

$\textcolor{b l u e}{\Delta E} = - {Z}^{2} \cdot \text{13.61 eV} \left(\frac{1}{3} ^ 2 - \frac{1}{1} ^ 2\right)$

$= - {3}^{2} \cdot \text{13.61 eV} \left(\frac{1}{9} - \frac{1}{1}\right)$

$=$ $\textcolor{b l u e}{\underline{\text{108.88 eV}}}$

And to check, we look on NIST. The energy levels for $n = 1 \to 3$ in ${\text{Li}}^{2 +}$ are listed as:

These are in ${\text{cm}}^{- 1}$. Take the first one for example (ignore the slight differences in the two choices).

${\lambda}_{r e f} = \frac{1}{\text{877919.12077 "cancel("cm"^(-1))) xx (10^7 "nm")/(cancel"1 cm}}$

$=$ $\text{11.3906 nm}$ $\approx$ $\text{11.39 nm}$

So, the wavelength we got is correct.