What is the wavelength of the photon emitted when an electron in a hydrogen atom which is in the initial state $n = 4$ jumps to the final state $n = 2$ ? $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } , 1 \mathrm { eV } = \right.$ $1.60 \times 10 ^ { - 19 } \mathrm {~J}$ )
A) $556 \mathrm {~nm}$
B) $312 \mathrm {~nm}$
C) $486 \mathrm {~nm}$
D) $243 \mathrm {~nm}$
E) $609 \mathrm {~nm}$

Question

Quizplus · Accepted Answer

The energy difference between two levels in a hydrogen atom is given by the formula $\Delta E = -13.6 \text{ eV} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where $n_i$ and $n_f$ are the initial and final principal quantum numbers, respectively. For a transition from $n = 4$ to $n = 2$, $\Delta E = -13.6 \text{ eV} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = -13.6 \text{ eV} \left( \frac{1}{4} - \frac{1}{16} \right) = -13.6 \text{ eV} \times \frac{3}{16} = -2.55 \text{ eV}$. Converting this energy to joules ($-2.55 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV}$) and then using the formula $E = \frac{hc}{\lambda}$ to find the wavelength $\lambda$, where $E$ is the energy in joules, $h$ is Planck's constant, and $c$ is the speed of light, we find that the wavelength corresponds to approximately $486 \text{ nm}$, which is the energy difference manifest as a photon's wavelength during the electron transition.

What Is the Wavelength of the Photon Emitted When an Electron