What is the number of de Broglie wavelengths | vedclass

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(A) According to Bohr's postulate for the quantization of angular momentum,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $

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$n$

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(A) According to Bohr's postulate for the quantization of angular momentum,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = mvr = \frac{nh}{2\pi}$.From the de Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting $mv = \frac{h}{\lambda}$ into the angular momentum equation,we get $\frac{h}{\lambda} \cdot r = \frac{nh}{2\pi}$.Rearranging the terms,we find $2\pi r = n\lambda$.Since $2\pi r$ is the circumference of the $n^{th}$ orbit,the equation $2\pi r = n\lambda$ implies that the circumference of the orbit is equal to $n$ times the de Broglie wavelength.Therefore,the number of de Broglie wavelengths in the $n^{th}$ orbit is $n$.

What is the number of de Broglie wavelengths associated with an electron revolving in the $n^{th}$ allowed orbit of a Bohr atom?

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