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If the De-Broglie wave length of the fourth Bohr’s orbit of hydrogen atom is $4 A^{0}$ , the circumference of the orbit is

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$4 \overset{o}{A}$

4 nm

$16 \overset{o}{A}$

16 nm

answer is C.

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Detailed Solution

$n = 4, λ = 4 \overset{°}{A} mvr = \frac{nh}{2 π} 2 πr = \frac{nh}{mv} 2 πr = nλ [∵ λ = \frac{h}{mv}] 2 πr = 4 \times 4 2 πr = 16 \overset{°}{A}$

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