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An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λ_n, λ_b be the de Broglie wavelength of the electron in the n^th state and the ground state respectively. Let Λ_n 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) Λ_n ≈ A + B/λ_n^2 (2) Λ_n ≈ A + B λ_n (3) Λ_n^2 ≈ A + B λ_n^2 (4) Λ_n^2 ≈ λ_n

1. Λ_n ≈ A + B/λ_n^2
2. Λ_n ≈ A + B λ_n
3. Λ_n^2 ≈ A + B λ_n^2
4. Λ_n^2 ≈ λ_n

Correct answer: Λ_n ≈ A + B/λ_n^2

Solution

The correct option indicates that the wavelength of the emitted photon, Λ_n, 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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