A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state. The atom returns to the ground state by emitting two photons, one in dropping to n = 4 state, and one in further dropping to the ground state. What are the photon wavelengths of (a) the first and (b) the second transitions?

Question

A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state. The atom returns to the ground state by emitting two photons, one in dropping to n = 4 state, and one in further dropping to the ground state. What are the photon wavelengths of (a) the first and (b) the second transitions?

Answer 1

To calculate the wavelengths of the photons emitted during the transitions, we can use the Rydberg formula, which relates the wavelength of an emitted photon to the initial and final energy levels of the transition.

The Rydberg formula is given by:

1/λ = R * (1/n_initial^2 - 1/n_final^2)

where λ represents the wavelength of the photon, R is the Rydberg constant (approximately 1.097 × 10^7 m^-1), and n_initial and n_final are the principal quantum numbers of the initial and final energy levels, respectively.

(a) First Transition: n = 5 to n = 4 state
For the first transition, the initial quantum number (n_initial) is 5, and the final quantum number (n_final) is 4.

Using the Rydberg formula:
1/λ1 = R * (1/5^2 - 1/4^2)

Simplifying the equation:
1/λ1 = R * (1/25 - 1/16)

Calculating:
1/λ1 = R * (16 - 25) / (25 * 16)
1/λ1 = R * (-9 / 400)

Simplifying further:
1/λ1 = -9R / 400
λ1 = -400 / (9R)

Substituting the value of R:
λ1 = -400 / (9 * 1.097 × 10^7)

Calculating:
λ1 ≈ -0.036 nm

The negative sign indicates that the photon is emitted, and the wavelength of the first transition is approximately 0.036 nm.

(b) Second Transition: n = 4 to ground state
For the second transition, the initial quantum number (n_initial) is 4, and the final quantum number (n_final) is 1 (ground state).

Using the Rydberg formula:
1/λ2 = R * (1/4^2 - 1/1^2)

Simplifying the equation:
1/λ2 = R * (1/16 - 1/1)

Calculating:
1/λ2 = R * (1 - 16) / 16
1/λ2 = R * (-15 / 16)

Simplifying further:
1/λ2 = -15R / 16
λ2 = -16 / (15R)

Substituting the value of R:
λ2 = -16 / (15 * 1.097 × 10^7)

Calculating:
λ2 ≈ -0.109 nm

Again, the negative sign indicates that the photon is emitted, and the wavelength of the second transition is approximately 0.109 nm.