If the radius of first orbit of H – atom is $a_{0}$ , then de- Broglie wavelength of electron in 4^th orbit is

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$2 π a_{0}$

$16 a_{0}$

$\frac{a_{0}}{4}$

$8 π a_{0}$

answer is A.

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Detailed Solution

$\begin{array}{l} r_{n} = a_{0} \times n^{2} \\ r_{4} = a_{0} \times {(4)}^{2} = 16 a_{0} \\ m v r = \frac{n h}{2 π}; m v = \frac{4 h}{2 π \times 16 a_{0}}; \\ λ = \frac{h}{m v} = \frac{h}{h / 8 π a_{0}} = 8 π a_{0} \end{array}$

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