If the radius of the first orbit of hydrogen atom is a₀, then de Broglie’s wave length of electron in 3rd orbit is

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3 ${πa}_{0}$

${πa}_{0}$ /6

${πa}_{0}$ /3

6 ${πa}_{0}$

answer is D.

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Detailed Solution

$\begin{array}{l} ℓ = \frac{0.529 \times n^{2}}{2} \\ ℓ_{1} = \frac{a_{0} \times (1)^{2}}{1} = a_{0} \\ ℓ_{3} = \frac{a_{0} \times 3^{2}}{1} = 9 a_{0} \\ 2 {πℓ}_{3} = nλ \\ 2 π 9 a_{0} = 3 λ \\ 6 {πa}_{0} = λ \end{array}$

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