A hydrogen atom makes a downward transition from the $n = 15$ state to the $n = 5$ state. Find the wavelength of the emitted photon. (c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19J, h = 6.626 × 10^-³⁴ J ∙ s)

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The energy of a hydrogen atom in the nth state is given by:$$E_n = -\frac{13.6 \text{ eV}}{n^2}$$The energy of the photon emitted when the atom makes a transition from the $n = 15$ state to the $n = 5$ state is:$$\Delta E = E_5 - E_{15} = -\frac{13.6 \text{ eV}}{5^2} - \left(-\frac{13.6 \text{ eV}}{15^2}\right) = 1.89 \text{ eV}$$The wavelength of the photon is given by:$$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{1.89 \text{ eV}(1.60 \times 10^{-19} \text{ J/eV})} = 2.56 \mu \text{m}$$

A Hydrogen Atom Makes a Downward Transition from The n=15n = 15n=15

A Hydrogen Atom Makes a Downward Transition from The $n = 15$