The de Broglie wavelength of an electron in t | vedclass

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

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equal to the circumference of the first orbit.

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(A) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $L = mvr = \frac{nh}{2\pi}$.Rearranging this,we get $2\pi r = n \frac{h}{mv}$.Since the de Broglie wavelength is defined as $\lambda = \frac{h}{mv}$,we can substitute this into the equation to get $2\pi r = n\lambda$.For the first orbit,$n = 1$,which gives $2\pi r = \lambda$.Therefore,the de Broglie wavelength is equal to the circumference of the first orbit.

The de Broglie wavelength of an electron in the first orbit of Bohr's model is . . . . . . .

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