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(A) 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

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$5$

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(A) 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).From the figure,$A$ corresponds to $n=2$,$B$ corresponds to $n=3$,and $C$ corresponds to $n=4$.The wavelength $\lambda$ for a transition from $n_2$ to $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For transition $\lambda_1$ (from $n=3$ to $n=2$): $\frac{1}{\lambda_1} = R(1)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left( \frac{5}{36} ight)$.For transition $\lambda_2$ (from $n=4$ to $n=3$): $\frac{1}{\lambda_2} = R(1)^2 \left[ \frac{1}{3^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{9} - \frac{1}{16} ight] = R \left( \frac{7}{144} ight)$.Now,calculate the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{R(7/144)}{R(5/36)} = \frac{7}{144} 	imes \frac{36}{5} = \frac{7}{4 	imes 5} = \frac{7}{20}$.Given the ratio is $\frac{7}{4n}$,we have $\frac{7}{20} = \frac{7}{4n}$,which implies $4n = 20$,so $n = 5$.

As per the given figure,$A$,$B$,and $C$ are the first,second,and third excited energy levels of a hydrogen atom,respectively. If the ratio of the two wavelengths (i.e.,$\frac{\lambda_1}{\lambda_2}$) is $\frac{7}{4n}$,then the value of $n$ will be

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