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(B) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \

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(B) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} ight)$,where $n = 3, 4, 5, \dots$ For the maximum wavelength (minimum energy),we take $n = 3$: $\frac{1}{\lambda_{max}} = 10^7 \left( \frac{1}{4} - \frac{1}{9} ight) = 10^7 \left( \frac{5}{36} ight) \implies \lambda_{max} = \frac{36}{5} 	imes 10^{-7} \ m = 7.2 	imes 10^{-7} \ m = 7200 \ Å$. For the minimum wavelength (maximum energy),we take $n = \infty$: $\frac{1}{\lambda_{min}} = 10^7 \left( \frac{1}{4} - 0 ight) = 10^7 \left( \frac{1}{4} ight) \implies \lambda_{min} = 4 	imes 10^{-7} \ m = 4000 \ Å$. The difference in wavelength is $\Delta \lambda = \lambda_{max} - \lambda_{min} = 7200 \ Å - 4000 \ Å = 3200 \ Å$.

The difference in the wavelength between the maximum and minimum of Balmer series [Use $R_{H} = 1 \times 10^7 \ m^{-1}$]. (in $Å$)

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