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(D) According to de Broglie's hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.According to Bohr's postulate for

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$\frac{2\pi r}{n}$

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(D) According to de Broglie's hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.According to Bohr's postulate for the quantization of angular momentum,$L = mvr_n = \frac{nh}{2\pi}$.Rearranging this equation to solve for the momentum $mv$,we get $mv = \frac{nh}{2\pi r_n}$.Substituting this expression for $mv$ into the de Broglie wavelength formula:$\lambda = \frac{h}{mv} = \frac{h}{(nh / 2\pi r_n)} = \frac{2\pi r_n}{n}$.Thus,the de Broglie wavelength of an electron in the $n^{	ext{th}}$ orbit is $\frac{2\pi r}{n}$.

For an electron moving in the $n^{\text{th}}$ Bohr orbit,the de Broglie wavelength of the electron is:

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