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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.For an electron in a hydrogen atom,the orb

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$\lambda_B = 3\lambda_G$

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(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.For an electron in a hydrogen atom,the orbital radius $r_n$ is proportional to $n^2$ and the velocity $v_n$ is proportional to $1/n$.The angular momentum quantization condition is $mvr = \frac{nh}{2\pi}$,which implies $2\pi r = n\lambda$.Thus,the de-Broglie wavelength $\lambda$ is proportional to the circumference of the orbit,$\lambda = \frac{2\pi r_n}{n}$.Since $r_n \propto n^2$,we have $\lambda \propto \frac{n^2}{n} = n$.For the ground state,$n_G = 1$,so $\lambda_G \propto 1$.For the second excited state,$n_B = 3$,so $\lambda_B \propto 3$.Therefore,$\frac{\lambda_B}{\lambda_G} = \frac{3}{1}$,which gives $\lambda_B = 3\lambda_G$.

The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of a hydrogen atom is related to that in the ground state $(\lambda_G)$ by:

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