The de-Broglie wavelength of an electron in t | vedclass

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(B) According to Bohr's quantization postulate,the angular momentum of an electron in a stationary orbit is given by:$mvr = \frac{nh}{2\pi}$Rearrangin

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$2 \pi r$

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(B) According to Bohr's quantization postulate,the angular momentum of an electron in a stationary orbit is given by:$mvr = \frac{nh}{2\pi}$Rearranging this equation,we get:$2\pi r = \frac{nh}{mv}$Since the de-Broglie wavelength is defined as $\lambda = \frac{h}{mv}$,we can substitute this into the equation:$2\pi r = n\lambda$For the ground state of the hydrogen atom,the principal quantum number is $n = 1$.Substituting $n = 1$ into the equation,we get:$\lambda = 2\pi r$

The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is

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