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Find the energy of hydrogen's electron in the level n = 3…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Find the energy of hydrogen's electron in the level $n = 3$ . Then find the wavelength and the frequency of the photon given out when it drops from $n = 3$ to $n = 2$ . Use $hc = 1240\ \text{eV nm}$ and $c = 3.0 \times 10^{8}\ \text{m/s}$ .

Need a reminder? The key ideaShow

Hydrogen's levels: $E_n = -\dfrac{13.6\ \text{eV}}{n^2}$ , and $-\dfrac{13.6\,Z^2}{n^2}\ \text{eV}$ for an ion with one electron. A jump gives a photon of energy $E_{\text{upper}} - E_{\text{lower}}$ , with $\lambda = \dfrac{1240\ \text{eV nm}}{E}$ . Lyman ends on $n = 1$ , Balmer on $n = 2$ , Paschen on $n = 3$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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