An electron in the $n=4$ level of the hydrogen atom relaxes to a lower energy level, emitting light of $\lambda=97.3 \, \text{nm}$.

Find the principal level to which the electron relaxed. Express your answer as an integer.

When an electron in a hydrogen atom relaxes from a higher energy level to a lower energy level, it emits light with a specific wavelength. In this case, the wavelength of the emitted light is 97.3 nm. The principal level to which the electron relaxed can be found using the formula:

ΔE = hc/λ

Where ΔE is the change in energy, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the emitted light.

Rearranging the formula to solve for n, the principal level, you get:

n = 1 - ΔE/RH

Where RH is the Rydberg constant (1.097 x 10^7 m^-1). Plugging in the values, you can calculate the principal level to be n = 2.

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An electron in the \(n=4\) level of the hydrogen atom relaxes to a lower energy level, emitting light of \(\lambda=97.3 \, \text{nm}\).Find the principal level to which the electron relaxed. Express your answer as an integer.

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An electron in the \(n=4\) level of the hydrogen atom relaxes to a lower energy level, emitting light of \(\lambda=97.3 \, \text{nm}\).

Find the principal level to which the electron relaxed. Express your answer as an integer.