When the excited electron in a hydrogen atom falls from n=5...

When the excited electron in a hydrogen atom falls from $n=5$ to $n=2$, a photon of blue light is emitted. If an excited electron in $\mathrm{He}^{+}$ falls from $n=4$, to which energy level must it fall so that a similar blue light (as with the hydrogen) is emitted? Prove it. (See Exercise 152.)

Key Concepts

Rydberg Formula

The Rydberg formula provides a way to calculate the wavelengths (or frequencies) of photons emitted during electronic transitions in hydrogen-like atoms. It encapsulates the relationship between the energy difference of two levels and the emitted photon’s properties, making it a crucial tool for understanding spectral lines and transitions in atomic systems.

Photon Emission from Electron Transitions

When an electron moves between energy levels within an atom, the difference in energy is emitted or absorbed as a photon. The energy (and thus the color) of the photon depends directly on the difference between the initial and final energy levels, linking the quantized nature of atomic energy levels to observable electromagnetic radiation.

Hydrogen-like Ions

Hydrogen-like ions are systems that consist of a nucleus and a single electron, such as hydrogen and He?. Although they follow the same energy level structure as the hydrogen atom, the presence of a different nuclear charge (Z) alters the energy values, making all electronic transitions more or less energetic by a factor of Z² compared to hydrogen.

Quantum Energy Levels

In quantum mechanics, electrons in an atom occupy discrete energy levels characterized by the principal quantum number, n. The energy associated with a specific level depends inversely on the square of n, and for hydrogen-like atoms, it also scales with the square of the nuclear charge (Z²), leading to an energy formula of the form E? ? -Z²/n².

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