An electron in the hydrogen atom (ground-state energy = -13.6 eV)makes a transition from the n = 2 to the n = 4 energy level.Calculate the magnitude of the energy of the photon involved in this process and state whether the photon was absorbed or emitted.
A)365 nm,absorbed
B)46 nm,emitted
C)7.6 nm,absorbed
D)1090 nm,absorbed
E)486 nm,absorbed

Question

Quizplus · Accepted Answer

The energy levels of a hydrogen atom are given by the formula $E_n = -\frac{13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. For a transition from $n = 2$ to $n = 4$, the energy difference is calculated as follows: $E_{initial} = -\frac{13.6}{2^2} = -3.4 \text{ eV}$ and $E_{final} = -\frac{13.6}{4^2} = -0.85 \text{ eV}$. The energy of the photon involved is $E_{photon} = E_{final} - E_{initial} = -0.85 - (-3.4) = 2.55 \text{ eV}$. Since the energy of the final state is higher than the initial state, the photon is absorbed. To find the wavelength, we use the relation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light. Rearranging for $\lambda$ gives $\lambda = \frac{hc}{E}$. Substituting $h = 4.135667696 \times 10^{-15} \text{ eV·s}$, $c = 3 \times 10^8 \text{ m/s}$, and $E = 2.55 \text{ eV}$, we find $\lambda \approx 486 \text{ nm}$, indicating that the photon was absorbed.

An Electron in the Hydrogen Atom (Ground-State Energy = -13