A hydrogen atom is in a quantum state with principal quantum number n = 9.The atom emits a photon and after emission its principal quantum number becomes 2.The wavelength of the emitted photon is
A)384 nm.
B)420 nm.
C)557 nm.
D)620 nm.
E)701 nm.

Question

Quizplus · Accepted Answer

The wavelength of the photon emitted when an electron transitions between two energy levels in a hydrogen atom can be calculated using the Rydberg formula: $ \frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) $, where $ \lambda $ is the wavelength of the emitted photon, $ R $ is the Rydberg constant ($1.097 \times 10^7 \, \text{m}^{-1}$), $ n_{1} $ is the lower energy level (2 in this case), and $ n_{2} $ is the higher energy level (9 in this case). Plugging in the values and solving for $ \lambda $ gives a wavelength in the vicinity of 384 nm.

A Hydrogen Atom Is in a Quantum State with Principal