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An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λₙ, λ_b be the de Broglie wavelength of the electron in the n^th state and the ground state respectively. Let Λₙ be the wavelength of the emitted photon in the transition from the n^th state to the ground state. For large n, (A, B are constants) (1) Λₙ ≈ A + B/λₙ² (2) Λₙ ≈ A + B λₙ (3) Λₙ² ≈ A + B λₙ² (4) Λₙ² ≈ λₙ

1. Λₙ ≈ A + B/λₙ²
2. Λₙ ≈ A + B λₙ
3. Λₙ² ≈ A + B λₙ²
4. Λₙ² ≈ λₙ

Correct answer: Λₙ ≈ A + B/λₙ²

Solution

The correct option indicates that the wavelength of the emitted photon, Λₙ, is inversely related to the square of the de Broglie wavelength of the electron in the n^th state, which aligns with the principles of quantum mechanics where higher energy transitions result in shorter wavelengths of emitted radiation.

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