The de-Broglie wavelength of the electron in | vedclass

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

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$3.33$

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(B) According to Bohr's quantization of angular momentum,the condition is given by:$mvr = \frac{nh}{2\pi}$Rearranging this,we get:$\frac{h}{mv} = \frac{2\pi r}{n} \quad \dots(i)$By definition,the de-Broglie wavelength $\lambda$ is given by:$\lambda = \frac{h}{mv} \quad \dots(ii)$Comparing equations $(i)$ and $(ii)$,we obtain:$\lambda = \frac{2\pi r}{n}$For the ground state of the hydrogen atom,$n = 1$ and the radius $r = 0.53 \ 	ext{Å}$.Substituting these values:$\lambda = \frac{2 	imes \pi 	imes 0.53 \ 	ext{Å}}{1}$$\lambda = 2 	imes 3.14159 	imes 0.53 \ 	ext{Å} \approx 3.33 \ 	ext{Å}$.

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

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