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(B) For the Lyman series, the shortest wavelength $(\lambda_0)$ corresponds to the transition from $n = \infty$ to $n = 1$.The energy of the photon is

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$6588$

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(B) For the Lyman series, the shortest wavelength $(\lambda_0)$ corresponds to the transition from $n = \infty$ to $n = 1$.The energy of the photon is given by $\frac{hc}{\lambda_0} = 13.6 	ext{ eV} 	imes \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = 13.6 	ext{ eV}$.Given $\lambda_0 = 915 	ext{ Å}$.For the Balmer series, the longest wavelength $(\lambda_1)$ corresponds to the transition from $n = 3$ to $n = 2$.The energy of the photon is given by $\frac{hc}{\lambda_1} = 13.6 	ext{ eV} 	imes \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 	ext{ eV} 	imes \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 	ext{ eV} 	imes \frac{5}{36}$.Dividing the two equations:$\frac{\lambda_1}{\lambda_0} = \frac{13.6}{13.6 	imes \frac{5}{36}} = \frac{36}{5}$.$\lambda_1 = \lambda_0 	imes \frac{36}{5} = 915 	imes \frac{36}{5} = 183 	imes 36 = 6588 	ext{ Å}$.

The shortest wavelength of the spectral lines in the Lyman series of the hydrogen spectrum is $915 \text{ Å}$. The longest wavelength of spectral lines in the Balmer series will be: (in $\text{ Å}$)

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