One of the emission spectral lines for Be^3+ has a...

One of the emission spectral lines for $\mathrm{Be}^{3+}$ has a wavelength of $253.4 \mathrm{~nm}$ for an electronic transition that begins in the state with $n=5 .$ What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one-electron ions. Don't forget the $Z$ factor: $Z=$ nuclear charge $=$ atomic number. $)$

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Bohr Model

The Bohr model describes electrons in atoms as orbiting the nucleus in fixed, quantized circular paths. In this model, each orbit corresponds to a specific energy level determined by the principal quantum number, and transitions between these orbits lead to the absorption or emission of electromagnetic radiation.

Hydrogen-like Ions

Hydrogen-like ions are ions that have been stripped down to one electron, similar to hydrogen, but with a nucleus that has a greater charge. In these ions, the energy levels are modified by the square of the nuclear charge (Z^2), meaning that the spacing between the energy levels is larger than in hydrogen.

Rydberg Formula

The Rydberg formula is used in the Bohr model to calculate the wavelengths of the spectral lines produced when an electron transitions between different energy levels. For hydrogen-like ions, the formula is adjusted to include the factor Z^2, accounting for the increased nuclear charge.

Principal Quantum Number

The principal quantum number (n) is a key quantum number that specifies the main energy levels or orbits of an electron in an atom. It is integral to determining the energy associated with each level in the Bohr model, and changes in n during an electron's transition between orbits result in the emission or absorption of radiation.

Transcript

Please give Ace some feedback