An electron in a Hydrogen atom first jumps fr | vedclass

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(D) The energy levels for a Hydrogen atom are given by $n=1$ (ground state),$n=2$ (first excited state),$n=3$ (second excited state),and $n=4$ (third

Vedclass · Accepted Answer

$\frac{20}{7}$

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(D) The energy levels for a Hydrogen atom are given by $n=1$ (ground state),$n=2$ (first excited state),$n=3$ (second excited state),and $n=4$ (third excited state).For the first transition from the third excited state $(n=4)$ to the second excited state $(n=3)$:$\frac{1}{\lambda_{1}} = R \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16-9}{144} ight) = R \left( \frac{7}{144} ight)$.For the second transition from the second excited state $(n=3)$ to the first excited state $(n=2)$:$\frac{1}{\lambda_{2}} = R \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = R \left( \frac{5}{36} ight)$.Taking the ratio $\frac{\lambda_{1}}{\lambda_{2}}$:$\frac{\lambda_{1}}{\lambda_{2}} = \frac{1/\lambda_{2}}{1/\lambda_{1}} = \frac{R(5/36)}{R(7/144)} = \frac{5}{36} 	imes \frac{144}{7} = \frac{5 	imes 4}{7} = \frac{20}{7}$.

An electron in a Hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_{1} : \lambda_{2}$ emitted in the two cases respectively is:

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