A hydrogen atom initially in the n = 6 state decays to the n = 2 state.The emitted photon is detected in a photographic plate.What is the wavelength of the detected photon? The lowest level energy state of hydrogen is -13.6 eV.(h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J, c = 3.00 × 10⁸ m/s) A)410 nm B)93.8 nm C)1090 nm D)93.1 nm

Question

Quizplus · Accepted Answer

The energy difference between the initial and final states is calculated using the formula for the energy levels of a hydrogen atom: $E_n = \frac{-13.6 \text{ eV}}{n^2}$. The energy of the photon emitted is $E = E_{initial} - E_{final} = \frac{-13.6 \text{ eV}}{6^2} - \frac{-13.6 \text{ eV}}{2^2}$. Converting this energy difference to joules and using $E = \frac{hc}{\lambda}$ allows solving for $\lambda$, the wavelength of the photon, which is approximately 410 nm.

A Hydrogen Atom Initially in the N = 6 State