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? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } \right.$ $h = 6.626 \times 10 ^{- 34} \mathrm {~J} \cdot \mathrm { s }$ $\left. 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} \right)$
A) 92 nm
B) 82 nm
C) 72 nm
D) 62 nm
E) 52 nm

Question

Quizplus · Accepted Answer

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

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

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