The de-Broglie wavelength of the electron in | vedclass

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

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$3.33$

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(B) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by:$mvr = \frac{nh}{2\pi}$Rearranging this,we get:$\frac{h}{mv} = \frac{2\pi r}{n}$ .......... $(i)$The de-Broglie wavelength $\lambda$ of an electron is given by:$\lambda = \frac{h}{mv}$ .......... $(ii)$Comparing equations $(i)$ and $(ii)$,we get:$\lambda = \frac{2\pi r}{n}$For the ground state (first orbit) of the hydrogen atom,$n = 1$ and $r = 0.53 \ \mathring{A}$.Substituting these values:$\lambda = \frac{2 	imes 3.1416 	imes 0.53}{1} \ \mathring{A} \approx 3.33 \ \mathring{A}$.

The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is . . . . . . $\mathring{A}$ (radius of the first orbit of hydrogen atom $= 0.53 \ \mathring{A}$).

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