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(C) According to Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.From the

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$2\pi r/n$

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(C) According to Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.From the de-Broglie relation, the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Rearranging the Bohr quantization condition, we get $mv = \frac{nh}{2\pi r}$.Substituting this into the de-Broglie wavelength formula: $\lambda = \frac{h}{(nh / 2\pi r)}$.Simplifying the expression, we get $\lambda = \frac{2\pi r}{n}$.

The de-Broglie wavelength of an electron moving in the $n^{th}$ Bohr orbit of radius $r$ is

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