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The de-Broglie wavelength of electron in Bohr orbits displays the following trend with increase in the value of quantum number n.

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Increase

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Decrease

answer is B.

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Detailed Solution

Since in Bohr theory $p = n (h / 2 π)$ , increase in p with increase in the value of n will be associated with decrease in de-Broglie wavelength. In fact, the expression of wavelength is $λ = 2 π / n$ .

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