If lambda_{1} and lambda_{2} are the waveleng | vedclass

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(A) According to the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.For an electron in the $n^{th}$

Vedclass · Accepted Answer

$1/2$

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(A) According to the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.For an electron in the $n^{th}$ Bohr orbit,the quantization condition is $mvr = \frac{nh}{2\pi}$,which implies $mv = \frac{nh}{2\pi r}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{nh / (2\pi r)} = \frac{2\pi r}{n}$.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.Thus,$\lambda_n = \frac{2\pi (a_0 n^2)}{n} = 2\pi a_0 n$.Therefore,the ratio of wavelengths for the first $(n=1)$ and second $(n=2)$ orbits is $\frac{\lambda_1}{\lambda_2} = \frac{2\pi a_0 (1)}{2\pi a_0 (2)} = \frac{1}{2}$.

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of de-Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom,then the ratio $\left(\frac{\lambda_{1}}{\lambda_{2}}\right)$ is equal to:

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