Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å ). The de-Broglie wavelength of this electron is:

Question

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å ). The de-Broglie wavelength of this electron is:, 12.9 Å, 6.6 Å, 9.7 Å, 3.5 Å

MARKS App · Accepted Answer

According to Bohr’s atomic model m v r = n h 2 π And the de-Broglie wavelength λ = h m v From the above two equations we can write, λ = 2 π r n In the 2 nd excited state λ = 2 π r 3 ∴ λ = 2 3 π × 4.65 Å = 9.7 Å

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