The ratio of the longest to shortest waveleng | vedclass

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Question

(A) For the Brackett series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2}

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$\frac{25}{9}$

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(A) For the Brackett series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$,where $n_1 = 4$ and $n_2 = 5, 6, 7, \dots$ The longest wavelength $(\lambda_{\max})$ corresponds to the transition from $n_2 = 5$ to $n_1 = 4$: $\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{4^2} - \frac{1}{5^2} ight] = R \left[ \frac{1}{16} - \frac{1}{25} ight] = R \left[ \frac{25 - 16}{400} ight] = \frac{9R}{400}$ $\lambda_{\max} = \frac{400}{9R}$ The shortest wavelength $(\lambda_{\min})$ corresponds to the transition from $n_2 = \infty$ to $n_1 = 4$: $\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{4^2} - \frac{1}{\infty^2} ight] = R \left[ \frac{1}{16} - 0 ight] = \frac{R}{16}$ $\lambda_{\min} = \frac{16}{R}$ The ratio of the longest to shortest wavelength is: $\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{400/9R}{16/R} = \frac{400}{9 	imes 16} = \frac{25}{9}$

The ratio of the longest to shortest wavelengths in the Brackett series of hydrogen spectra is

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