Calculate the wavelength of light emitted when each of the following transactions occur in the hydrogen atom.what type of electromagnetic radiation is emitted in each transion n=5-n=3
The transition from n=5 to n=3 in hydrogen atom corresponds to the emission of a photon. The energy difference between the initial and final states is given by the Rydberg formula:
1/λ = R∞ (1/n1^2 - 1/n2^2)
where λ is the wavelength of the emitted photon, R∞ is the Rydberg constant, and n1 and n2 are the initial and final quantum numbers, respectively.
For this transition, n1 = 5 and n2 = 3. Substituting these values into the formula, we get:
1/λ = R∞ (1/5^2 - 1/3^2)
1/λ = R∞ (1/25 - 1/9)
1/λ = R∞ (16/225)
Solving for λ, we get:
λ = 225/(16 R∞)
At room temperature, R∞ = 1.097 x 10^7 m^-1. Substituting this value, we get:
λ = 225/(16 x 1.097 x 10^7)
λ = 1.213 x 10^-6 m
Therefore, the wavelength of the emitted radiation is 1.213 x 10^-6 m, which corresponds to the infrared region of the electromagnetic spectrum.