Question

in a hypothetical atom, match the electron transition to the color of light emitted. red choose choose yellow n=6 --> n=2 n=5 --> n=4 n=5 --> n=3 violet n=3 --> n=4

Explanation:

The energy of a photon emitted during an electron - transition is given by $E = h
u=\Delta E$, where $\Delta E$ is the energy difference between the two levels. The energy difference $\Delta E$ is proportional to the frequency $
u$ of the emitted light. Higher - energy transitions correspond to higher - frequency (shorter - wavelength) light. The energy levels in an atom are quantized, and the energy difference between levels is larger for transitions involving higher - numbered levels. The transition $n = 6
ightarrow n = 2$ has a large energy difference and corresponds to violet light. The transition $n=5
ightarrow n = 3$ has a moderate energy difference and corresponds to yellow light. The transition $n = 5
ightarrow n = 4$ has a small energy difference and corresponds to red light. The transition $n = 3
ightarrow n = 4$ is an absorption process (not an emission as required here, so we ignore it for the purpose of matching emissions).

Answer:

Red: $n = 5
ightarrow n = 4$
Yellow: $n = 5
ightarrow n = 3$
Violet: $n = 6
ightarrow n = 2$