The shortest wavelength in the hydrogen spect | vedclass

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(B) The spectral series in the hydrogen spectrum are Lyman $(n_1=1)$,Balmer $(n_1=2)$,Paschen $(n_1=3)$,Brackett $(n_1=4)$,and Pfund $(n_1=5)$.For the

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$91.2$

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(B) The spectral series in the hydrogen spectrum are Lyman $(n_1=1)$,Balmer $(n_1=2)$,Paschen $(n_1=3)$,Brackett $(n_1=4)$,and Pfund $(n_1=5)$.For the shortest wavelength,the energy transition must be maximum,which occurs for the Lyman series $(n_1=1)$ with $n_2=\infty$.Using the Rydberg formula: $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.Substituting $n_1=1$ and $n_2=\infty$: $\frac{1}{\lambda} = R_H \left[ \frac{1}{1^2} - \frac{1}{\infty^2} ight] = R_H$.Given $R_H \approx 1.097 	imes 10^7 \ m^{-1}$,we get $\lambda = \frac{1}{R_H} \approx 9.117 	imes 10^{-8} \ m = 91.2 \ nm$.

The shortest wavelength in the hydrogen spectrum is approximately (in $nm$)

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