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In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, $a_{0} =52.9 pm$ ]

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$211.6 p m$

$105.3 pm$

$211.6 π pm$

$52.9 π pm$

answer is B.

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Detailed Solution

$nλ=2πr$

${nλ=2πr}_{H} {.n}^{2}$

${λ=2πr}_{H} .n=2 \times π \times 52.9 \times 2=211.6 π$

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