The number of de-Broglie wavelengths containe | vedclass

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(B) According to Bohr's quantization condition, the angular momentum of an electron in an orbit is given by $L = mvr = \frac{nh}{2\pi}$.From the de-Br

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$2$

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(B) According to Bohr's quantization condition, the angular momentum of an electron in an orbit is given by $L = mvr = \frac{nh}{2\pi}$.From the de-Broglie hypothesis, the wavelength of an electron is $\lambda = \frac{h}{mv}$.Substituting $mv = \frac{h}{\lambda}$ into the quantization condition, we get $\frac{h}{\lambda} r = \frac{nh}{2\pi}$.This simplifies to $2\pi r = n\lambda$, where $2\pi r$ is the circumference of the $n^{th}$ orbit.For the second Bohr orbit, $n = 2$.Therefore, the circumference of the second orbit is $2\pi r = 2\lambda$.This implies that the number of de-Broglie wavelengths contained in the second Bohr orbit is $2$.

The number of de-Broglie wavelengths contained in the second Bohr orbit of a hydrogen atom is

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