An electron in the ground state of the hydrog | vedclass

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.According to Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$,which implies $m

Vedclass · Accepted Answer

$1/4$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.According to Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$,which implies $mv = \frac{nh}{2\pi r}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{nh / (2\pi r)} = \frac{2\pi r}{n}$.Thus,$\lambda \propto \frac{r}{n}$.For the ground state,$n_1 = 1$ and $r_1 = 5.3 	imes 10^{-11} \ m$.For the third excited state,$n_4 = 4$ and $r_4 = 8.48 	imes 10^{-10} \ m = 84.8 	imes 10^{-11} \ m$.The ratio is $\frac{\lambda_1}{\lambda_4} = \frac{r_1}{n_1} 	imes \frac{n_4}{r_4} = \frac{5.3 	imes 10^{-11}}{1} 	imes \frac{4}{84.8 	imes 10^{-11}} = \frac{5.3 	imes 4}{84.8} = \frac{21.2}{84.8} = \frac{1}{4}$.

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