Question

An excited Hydrogen atom in the n=3 energy level decays to the n=2 energy level

and emits a photon of light. If the energies of these two levels are separated by
3.029 x 10⁻¹⁹ J, what will be the wavelength of the light emitted in nanometers?

Answer 1

The wavelength of light emitted when an excited hydrogen atom in the n=3 energy level decays to the n=2 energy level can be calculated using the equation λ = hc/E, where λ is the wavelength, h is Planck's constant, c is the speed of light, and E is the energy difference.

Step-by-step explanation:

The energy difference between two energy levels can be related to the wavelength of light emitted using the equation:

E = hc/λ

Where E is the energy difference, 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 light. Rearranging the equation, we get:

λ = hc/E

Using the given energy difference of 3.029 x 10^-19 J, we can substitute the values into the equation to calculate the wavelength:

λ = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/(3.029 x 10^-19 J)

Simplifying the calculation, we find that the wavelength is approximately 656 nm.