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(D) The shortest wavelength $(\lambda_{\min})$ in the entire hydrogen emission spectrum occurs for the transition from $n = \infty$ to $n = 1$ (Lyman

Vedclass · Accepted Answer

$900/11$

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(D) The shortest wavelength $(\lambda_{\min})$ in the entire hydrogen emission spectrum occurs for the transition from $n = \infty$ to $n = 1$ (Lyman series limit).Using the Rydberg formula: $\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.Thus, $\lambda_{\min} = \frac{1}{R}$.The longest wavelength $(\lambda_{\max})$ among the five series (Lyman, Balmer, Paschen, Brackett, Pfund) occurs for the transition from $n = 6$ to $n = 5$ (the last line of the Pfund series).Using the Rydberg formula: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{5^2} - \frac{1}{6^2} \right) = R \left( \frac{1}{25} - \frac{1}{36} \right) = R \left( \frac{36 - 25}{900} \right) = \frac{11R}{900}$.Thus, $\lambda_{\max} = \frac{900}{11R}$.The ratio of the longest wavelength to the shortest wavelength is $\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{900/11R}{1/R} = \frac{900}{11}$.

The ratio of the longest wavelength to the shortest wavelength observed in the five spectral series of the emission spectrum of hydrogen is

Similar Questions

Assertion : In Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason : Lyman series constitute spectral lines corresponding to transition from higher energy to ground state of hydrogen atom.

The spectral series of the hydrogen atom that lies in the visible region of the electromagnetic spectrum is:

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the first line of the Lyman and Paschen series respectively,then $\lambda_2 : \lambda_1$ is

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the third member of the Lyman series and the first member of the Paschen series respectively,then the value of $\lambda_{1} : \lambda_{2}$ is

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