For the Bohr's first orbit of circumference 2 | vedclass

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(A) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this eq

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$2\pi r$

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(A) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this equation,we get $2\pi r = n \left( \frac{h}{mv} \right)$.From the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting this into the Bohr's condition,we get $2\pi r = n\lambda$.For the first orbit,$n = 1$.Therefore,$2\pi r = 1 \cdot \lambda$,which implies $\lambda = 2\pi r$.

For the Bohr's first orbit of circumference $2\pi r$,the de-Broglie wavelength of the revolving electron will be

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