The longest wavelength of a photon that can be emitted by a hydrogen atom, for which the initial state is $n = 3$, is closest to which one of the following values? $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \right.$ $\mathrm { s } , 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J}$ )
A) $625 \mathrm {~nm}$
B) $600 \mathrm {~nm}$
C) $658 \mathrm {~nm}$
D) $575 \mathrm {~nm}$
E) $550 \mathrm {~nm}$

Question

Quizplus · Accepted Answer

The longest wavelength corresponds to the smallest energy transition, which is from $n=3$ to $n=2$. Using the Rydberg formula for hydrogen, $\frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)$, where $R_H = 1.097 \times 10^7 \, \text{m}^{-1}$, we find $\lambda \approx 656 \, \text{nm}$, which is closest to 658 nm.

The Longest Wavelength of a Photon That Can Be Emitted n=3n = 3n=3

The Longest Wavelength of a Photon That Can Be Emitted $n = 3$