If lambda_1 and lambda_2 denote the wavelengt | vedclass

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(B) According to Bohr's quantization condition,the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.Rearrangi

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$1/2$

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(B) According to Bohr's quantization condition,the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this,we get $2\pi r = \frac{nh}{mv}$.Since the de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$,the condition becomes $2\pi r = n\lambda$,or $\lambda = \frac{2\pi r}{n}$.For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n = a_0 n^2$,where $a_0$ is the Bohr radius.Substituting $r_n$ into the wavelength formula: $\lambda_n = \frac{2\pi (a_0 n^2)}{n} = 2\pi a_0 n$.For the first orbit $(n=1)$,$\lambda_1 = 2\pi a_0 (1) = 2\pi a_0$.For the second orbit $(n=2)$,$\lambda_2 = 2\pi a_0 (2) = 4\pi a_0$.Therefore,the ratio $\frac{\lambda_1}{\lambda_2} = \frac{2\pi a_0}{4\pi a_0} = \frac{1}{2}$.

If $\lambda_1$ and $\lambda_2$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom,then $\lambda_1/\lambda_2$ is equal to

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