Which electron energy level transition corresponds to a hydrogen atom absorbing a visible light photon that has a wavelength of 656 nanometers?

Question: Which electron energy level transition corresponds to a hydrogen atom absorbing a visible light photon that has a wavelength of 656 nanometers?

The electron energy level transition in a hydrogen atom that corresponds to absorbing a visible light photon with a wavelength of 656 nanometers is from the second energy level to the third energy level.

This can be calculated using the following equation:

E = hc / λ

where:

• E is the energy of the photon in joules
• h is Planck's constant (6.62607015 × 10⁻³⁴ J⋅s)
• c is the speed of light (299,792,458 m/s)
• λ is the wavelength of the photon in meters

Plugging in the values for the wavelength of 656 nanometers, we get:

E = (6.62607015 × 10⁻³⁴ J⋅s) (299,792,458 m/s) / (656 × 10⁻⁹ m)

E = 3.03 × 10⁻¹⁹ J

This energy corresponds to the energy difference between the second and third energy levels in a hydrogen atom. Therefore, the electron energy level transition that corresponds to absorbing a visible light photon with a wavelength of 656 nanometers is from the second energy level to the third energy level.

This transition is also known as the Balmer transition, and it is responsible for the red color in the visible spectrum of hydrogen.

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