What is the wavelength of the photon emitted when a hydrogen atom which is in the initial state n = 4 transitions to the final state n = 2? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J) A)243 nm B)486 nm C)556 nm D)312 nm E)609 nm

Question

Quizplus · Accepted Answer

The energy difference between two levels in a hydrogen atom is given by the formula $\Delta E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the emitted photon. The energy levels of a hydrogen atom can also be calculated using the formula $E_n = -\frac{13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. The energy difference between the initial state $n = 4$ and the final state $n = 2$ is $\Delta E = E_2 - E_4 = -\frac{13.6 \text{ eV}}{2^2} - \left(-\frac{13.6 \text{ eV}}{4^2}\right)$. Calculating this gives $\Delta E = 2.55 \text{ eV}$, which can be converted to joules ($1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$). Substituting $\Delta E$, $h$, and $c$ into the first formula allows solving for $\lambda$, which results in approximately 486 nm.

What Is the Wavelength of the Photon Emitted When a Hydrogen