The de-Broglie's wavelength of an electron pr | vedclass

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(B) The radius of the first Bohr orbit of an $H$ atom is given by $r = 0.529 \ \mathring{A}$.According to Bohr's postulate,the angular momentum is qua

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$2\pi 	imes 0.529 \ \mathring{A}$

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(B) The radius of the first Bohr orbit of an $H$ atom is given by $r = 0.529 \ \mathring{A}$.According to Bohr's postulate,the angular momentum is quantized as $mvr = \frac{nh}{2\pi}$.For the first orbit,$n = 1$,so $mvr = \frac{h}{2\pi}$.Rearranging this gives $2\pi r = \frac{h}{mv}$.According to the de-Broglie relation,$\lambda = \frac{h}{mv}$.Therefore,$\lambda = 2\pi r$.Substituting the value of $r$,we get $\lambda = 2\pi 	imes 0.529 \ \mathring{A}$.

The de-Broglie's wavelength of an electron present in the first Bohr orbit of an $H$ atom is:

Similar Questions

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