Question

For hydrogen, what is the wavelength of the photon emitted when an electron drops from a 4p orbital to a 2s orbital in a hydrogen atom?

Answer 1

In Bohr's atomic model, the electrons are orbiting outside in orbitals around the nucleus. The farther the electron is from the nucleus, the lower its energy level becomes. That is why when reactions occur, it is the valence electrons (outermost electrons) that gets involve in the bonding. The way you write an electronic configuration is how the energy levels decreases. The first is orbital 1s which is the highest energy level because it is nearest to the nucleus. Then, it is followed by 2s2p, and so on and so forth. The energy levels are represented by the numbers.

When electrons transfer from orbital to orbital, they may release (high to low) or absorb (low to high) energy in the form of light which can be measuredin wavelength. The formula to be used is Rydberg's formula:

1/λ = R(1/n₁² - 1/n₂²), where

λ is wavelength measured in meters
n₁ and n₂ are the energy levels such that n₂>n₁
R is the Rydberg constant equal to 1.097×10⁷ m⁻¹

1/λ =1.097×10⁷ m⁻¹ (1/2² - 1/4²)
λ = 4.86×10⁻⁷ or 4.86 pm