The de-broglie wavelength of an electron in the 3rd orbit of hydrogen atom is

9.97 A^{0}

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8.0 A^{0}

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10.20 A^{0}

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12.80 A^{0}

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Solution

The correct option is A $9.97 A^{0}$
Radius of $3 r d$ orbit $= r = R (3)^{2}$
$= 9 R$
$[$ Here $R =$ Radius of first orbit of hydrogen atom $]$
$m v r = n h / (2 π)$
$h / (m v) = 2 π r / n$
$λ = h / (m v) = 2 π \times 9 R / 3$
$λ = 6 π R$
$λ = 6 π \times (0.529 Å)$
$λ = 9.97 Å$
De-Broglie wavelength is $9.97 Å$
Hence,
option $(A)$ is correct answer.

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