The longest wavelength photon that can be emitted by a hydrogen atom,for which the final state is $n = 9 ,$ is closest to which one of the following values? (c = 3.0 × 10⁸ m/s, $1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J}$ $\left. h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$ A)39,000 nm B)22,000 nm C)7400 nm D)16,000 nm E)28,000 nm

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The longest wavelength photon corresponds to the smallest energy difference between the levels involved. The energy difference between levels $n_i$ and $n_f$ is given by:$$\Delta E = E_i - E_f = -13.6 \text{ eV} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$For the longest wavelength photon, $n_i = \infty$ and $n_f = 9$. Plugging these values into the equation, we get:$$\Delta E = -13.6 \text{ eV} \left(\frac{1}{9^2} - \frac{1}{\infty^2}\right) = -13.6 \text{ eV} \left(\frac{1}{81}\right) = 0.168 \text{ eV}$$Using the relationship between energy and wavelength:$$\lambda = \frac{hc}{\Delta E}$$we get:$$\lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.0 \times 10^8 \text{ m/s})}{0.168 \times 1.60 \times 10^{-19} \text{ J}} = 39,000 \text{ nm}$$Therefore, the longest wavelength photon that can be emitted by a hydrogen atom, for which the final state is $n = 9$, is closest to 39,000 nm.

The Longest Wavelength Photon That Can Be Emitted by a Hydrogen