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(B) According to the de-Broglie hypothesis,the wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,

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

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(B) According to the de-Broglie hypothesis,the wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity of the electron. Rearranging this,we get $mv = \frac{h}{\lambda} \quad --- (1)$. According to Bohr's postulate for the quantization of angular momentum,the angular momentum $L$ of an electron in the $n^{	ext{th}}$ orbit is given by $L = mvr = \frac{nh}{2\pi}$. Rearranging this for $mv$,we get $mv = \frac{nh}{2\pi r} \quad --- (2)$. Equating the expressions for $mv$ from equation $(1)$ and $(2)$: $\frac{h}{\lambda} = \frac{nh}{2\pi r}$. Canceling $h$ from both sides,we get $\frac{1}{\lambda} = \frac{n}{2\pi r}$. Therefore,the de-Broglie wavelength is $\lambda = \frac{2\pi r}{n}$.

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

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