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Q.
43. For hydrogen atom electron in $ {{text{n}}^{text{th}}} $ Bohr orbit, the ratio of radius of orbit to its de-Broglie wavelength is – LIVE ANSWER TODAYSkip to content

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43. For hydrogen atom electron in $ {{text{n}}^{text{th}}} $ Bohr orbit, the ratio of radius of orbit to its de-Broglie wavelength is

Solution:

For nth Bohr orbit, $ r=frac{{{varepsilon }_{0}}{{n}^{2}}{{h}^{2}}}{pi mZ{{e}^{2}}} $ de-Broglie wavelength $ lambda =frac{h}{mv} $ Ratio of both r and $ lambda $ , we have $ frac{r}{lambda }=frac{{{varepsilon }_{0}}{{n}^{2}}{{h}^{2}}}{pi mZ{{e}^{2}}}times frac{mv}{h} $ $ =frac{{{varepsilon }_{0}}{{n}^{2}}hv}{pi Z{{e}^{2}}} $ But $ v=frac{Z{{e}^{2}}}{2h{{varepsilon }_{0}}n} $ for nth orbit Hence, $ frac{r}{lambda }=frac{n}{2pi } $