An electron in the ground state of the hydrogen atom has the orbital radius of $5.3\times {10}^{-11}\mathrm{m}$ while that for the electron in third excited state is $8.48\times {10}^{-10}\mathrm{m}$. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is:

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Correct answer: b) $\frac{1}{4}$. $\text{ }\lambda =\frac{h}{mv}\ mvr=\frac{nh}{2\pi },\lambda =\frac{2\pi rh}{nh}\ \lambda \propto \frac{r}{n}\ \frac{{\lambda }_{1}}{{\lambda }_{4}}=\frac{{r}_{1}{n}_{4}}{{n}_{1}{r}_{4}}=\frac{5.3\times {10}^{-11}\times 4}{1\times 84.8\times {10}^{-11}}\ =\frac{1}{4}$

An electron in the ground state of the hydrogen atom has the orbital radius of \(5.3\times {10}^{-11}\mathrm{m}\) while …

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