How do you calculate the wavelength of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level? Recall that for hydrogen $E_n=-2.18\times {10}^{-18}J\left(\frac{1}{n^2}\right)$

Note that you were given only one energy state. If you consider two energy states, from $n=4$ to $n=1$, we have:

$E_1-E_4=\Delta E$

$=-2.18\times {10}^{-18}\text{J}(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

$=-2.18\times {10}^{-18}\text{J}(\frac{1}{1^2}-\frac{1}{4^2})$

$=-2.18\times {10}^{-18}\text{J}\left(\frac{15}{16}\right)$

$=$ $-2.04\times {10}^{-18}\text{J}$

After you obtain the energy, then you can realize that that energy has to correspond exactly to the energy of the photon that came in:

$|\Delta E|=E_{\text{photon}}=h\nu =\frac{hc}{\lambda }$

where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incoming photon. Thus, the wavelength is:

$\Rightarrow \lambda =\frac{hc}{E_{\text{photon}}}=\frac{(6.626\times {10}^{-34}\text{J}\cdot \text{s})(2.998\times {10}^8\text{m/s})}{2.04\times {10}^{-18}\text{J}}$

$=9.720\times {10}^{-8}$ $\text{m}$

$=$ $\text{97.20 nm}$