Problem 2:

In a hydrogen atom, what is the wavelength of the photon emitted when an electron transitions from:

(a) The $ n = 4 $ state to the $ n = 2 $ state.
Answer: 487.1 nm.

(b) The $ n = 4 $ state to the $ n = 1 $ state.
Answer: 97.4 nm.

(c) The $ n = 5 $ state to the $ n = 2 $ state.
Answer: 434.3 nm.

Question

26-09-2024
Business

High School

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Problem 2:

In a hydrogen atom, what is the wavelength of the photon emitted when an electron transitions from:

(a) The $ n = 4 $ state to the $ n = 2 $ state.
Answer: 487.1 nm.

(b) The $ n = 4 $ state to the $ n = 1 $ state.
Answer: 97.4 nm.

(c) The $ n = 5 $ state to the $ n = 2 $ state.
Answer: 434.3 nm.

Answer :

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jaiswaljyoti jaiswaljyoti · Answer 1 · 2024-03-31T08:29:15-04:00

In a hydrogen atom, the wavelength of the photon emitted during electron transitions can be calculated using the Rydberg formula.

The wavelength of the photon emitted during a transition in a hydrogen atom can be determined using the Rydberg formula:

1/λ = R_H * (1/n_f^2 - 1/n_i^2)

Where λ represents the wavelength, R_H is the Rydberg constant (approximately 1.097 × 10^7 m^-1), and n_f and n_i are the final and initial energy levels, respectively.

(a) For the transition from n = 4 to n = 2:

1/λ = 1.097 × 10^7 * (1/2^2 - 1/4^2)

Solving this equation gives us:

λ = 487.1 nm

(b) For the transition from n = 4 to n = 1:

1/λ = 1.097 × 10^7 * (1/1^2 - 1/4^2)

Solving this equation gives us:

λ = 97.4 nm

(c) For the transition from n = 5 to n = 2:

1/λ = 1.097 × 10^7 * (1/2^2 - 1/5^2)

Solving this equation gives us:

λ = 434.3 nm

These calculations are based on the principles of quantum mechanics and the energy differences between the electron orbitals in the hydrogen atom. The emitted photon's wavelength corresponds to the energy difference between the initial and final states of the electron transition.

To learn more about the Rydberg formula click here: brainly.com/question/13185515

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