The ratio of maximum to minimum wavelength in Balmer series of hydrogen atom is
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You can find this ratio directly without full fraction inversions by dividing the brackets inside the Rydberg setup: $\frac{\lambda_{max}}{\lambda_{min}} = \frac{(1/4 - 1/\infty)}{(1/4 - 1/9)} = \frac{1/4}{5/36} = \frac{36}{20} = \frac{9}{5}$. This layout saves several algebra transformation steps.
Step 1: Translate wavelength extremes into energy extremes. Longest wavelength means smallest energy jump; shortest wavelength means largest jump. In the Balmer series the lower level is fixed at $n_1 = 2$. Step 2: Longest wavelength transition. The smallest jump comes from the nearest level, $n_2 = 3$: $\dfrac{1}{\lambda_{\max}} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = R\cdot\dfrac{5}{36}$. Step 3: Shortest wavelength transition. The largest jump is from $n_2 = \infty$: $\dfrac{1}{\lambda_{\min}} = R\left(\dfrac{1}{4} - 0\right) = \dfrac{R}{4}$. Step 4: Form the wavelength ratio. $\dfrac{\lambda_{\max}}{\lambda_{\min}} = \dfrac{1/\lambda_{\min}}{1/\lambda_{\max}} = \dfrac{R/4}{5R/36}$. Step 5: Simplify. $= \dfrac{1}{4}\times\dfrac{36}{5} = \dfrac{36}{20} = \dfrac{9}{5}$. Step 6: Conclude. The ratio of maximum to minimum wavelength is $9:5$. \[ \boxed{\lambda_{\max} : \lambda_{\min} = 9 : 5} \]