# What happens when you give excess energy to an atom?

So this is my question:

An electron in a hydrogen atom in its ground state absorbs energy equal to the ionisation energy of $Li^{2+}$. The wavelength of the emitted electron is?

I started off by finding out the Ionisation energy of $Li^{2+}$:

$$I.E = 13.6 \bigg(\frac{z^2}{n^2}\bigg) = 9(13.6)$$

When this energy is given to a Hydrogen atom in the ground state, it'll get ionised by using up energy $(=13.6 eV)$ and the rest of the energy $(=8\small(13.6\small)eV)$ will be its kinetic energy. I don't know how to proceed further. I tried: $$E =\frac{hc}{\lambda}$$ but that doesn't give me an answer. Any help?

Something that's confusing me is that in all previous questions, we've talked about the wavelength of the photon required to change the state of an electron. Here, we're giving energy to an atom and then we're supposed to find out the wavelength of that ionised electron itself.

Your question is asking you to calculate the de Broglie wavelength of the electron, which can be obtained from its momentum $p$ via de Broglie's relation $$\lambda=\frac h p$$ where $h$ is Planck's constant.

Regarding units:

You should work consistently with your units. You can work with the kinetic energy in Joules, $h$ in J s and $p$ in kg m/s. But you can also work with $E$ in hartrees, $p$ in atomic units, $hbar=1$, and get $λ$ in multiples of the Bohr radius.

The formula you mentioned in the comments, $$\lambda = \sqrt{\frac{150}{\text{K.E. in }e\text V}}\text Å,$$ works too, though it has limited precision. I would encourage you instead to never drop the units from your calculation: $$\lambda=\frac{h}{\sqrt{2m K}}=\frac{6.6\times10^{-34}\,\text{kg}\,\text m^2\,\text s^{-1}}{\sqrt{2\times9.1\times10^{-31}\,\text{kg}\times 8\times13.6\,e\text V}} =4.7\times10^{-20}\frac{\text{kg}\,\text m^2\,\text s^{-1}}{\sqrt{\text{kg}\,e\text V}},$$ and since $1\,e\text V=1.6\times10^{-19}\,\text J$, you have $$1\frac{\text{kg}\,\text m^2\,\text s^{-1}}{\sqrt{\text{kg}\,e\text V}} =\frac{1}{\sqrt{1.6\times10^{-19}}}\frac{\text{kg}\,\text m^2\,\text s^{-1}}{\sqrt{\text{kg}^2\,\text m^2\,\text{s}^{-2}}}=2.5\times10^9\,\text m,$$ which gives $$\lambda=1.2\times10^{-10}\,\text m=1.2\,\text Å.$$

• Okay, so $p = \sqrt{2mK}$, ($K = 8(13.6) = 108.8$). I substituted the values, but didn't get an answer which matches :/ – mikhailcazi Dec 2 '13 at 11:24
• <lightbulb!> Should I have converted the KE value to joules? – mikhailcazi Dec 2 '13 at 11:30
• You should work consistently with your units. You can work with the kinetic energy in Joules, $h$ in Js and $p$ in kg m/s. But you can also work with $E$ in hartrees, $p$ in atomic units, $\hbar=1$, and get $\lambda$ in multiples of the Bohr radius. – Emilio Pisanty Dec 2 '13 at 11:37
• Well, I took mass in $kg$, Planck's constant in $m^2 kg/s$, and need $\lambda$ in metres (or $\mu m$ or Angstrom). – mikhailcazi Dec 2 '13 at 12:02
• Then you should work completely in SI units, of course. – Emilio Pisanty Dec 2 '13 at 12:04