The de Broglie wavelength of an electron revo | vedclass

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(A) According to Bohr's postulate,the angular momentum of an electron is given by $mvr = nh/2\pi$.For the first orbit,$n = 1$,so $mvr = h/2\pi$.Rearra

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$2\pi r$

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(A) According to Bohr's postulate,the angular momentum of an electron is given by $mvr = nh/2\pi$.For the first orbit,$n = 1$,so $mvr = h/2\pi$.Rearranging this,we get $mv = h/(2\pi r)$.The de Broglie wavelength $\lambda$ is defined as $\lambda = h/p = h/(mv)$.Substituting the value of $mv$,we get $\lambda = h / (h / 2\pi r) = 2\pi r$.Thus,the de Broglie wavelength of the electron in the first Bohr orbit is equal to the circumference of the orbit,which is $2\pi r$.

The de Broglie wavelength of an electron revolving in the first Bohr orbit of circumference $2\pi r$ is ............

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