Question
The de-Broglie wavelength $({\lambda _B})$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $({\lambda _G})$ by
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Answer
de-Broglie wavelength, $\lambda=\frac{h}{P}$
$\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{G}}}=\frac{\mathrm{P}_{\mathrm{G}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\mathrm{mv}_{\mathrm{G}}}{\mathrm{mv}_{\mathrm{B}}}$
Speed of electron $v \propto \frac{z}{n}$
so $\frac{\lambda_{B}}{\lambda_{G}}=\frac{n_{B}}{n_{G}}=\frac{3}{1} \Rightarrow \lambda_{B}=3 \lambda_{G}$
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