The shortest wavelength of a photon that can be emitted by a hydrogen atom, for which the initial state is $n = 12$ is closest to which one of the following values? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \right.$ $h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } ,$ $\left. 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} \right)$

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Quizplus · Accepted Answer

The shortest wavelength corresponds to the largest energy difference between the initial and final states. The energy difference between the initial state $n = 12$ and the ground state $n = 1$ is given by:$$\Delta E = E_f - E_i = -13.6 \text{ eV}\left(\frac{1}{1^2} - \frac{1}{12^2}\right) = 13.59 \text{ eV}$$Converting this energy to wavelength using the relation $E = hc/\lambda$, we get:$$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \text{ J s})(3.0 \times 10^8 \text{ m/s})}{13.59 \text{ eV}(1.60 \times 10^{-19} \text{ J/eV})} = 92.1 \text{ nm}$$Therefore, the shortest wavelength is closest to 92 nm, which is choice A.

The Shortest Wavelength of a Photon That Can Be Emitted n=12n = 12n=12

The Shortest Wavelength of a Photon That Can Be Emitted $n = 12$