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The ratio of the largest to shortest wav...

The ratio of the largest to shortest wavelength in Balmer series of hydrogen spectra is,

A

`(25)/(9)`

B

`(17)/(6)`

C

`(9)/(5)`

D

`(5)/(4)`

Text Solution

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The correct Answer is:
To find the ratio of the largest to shortest wavelength in the Balmer series of hydrogen spectra, we can follow these steps: ### Step 1: Understand the Rydberg Formula The wavelength (\( \lambda \)) for the Balmer series can be expressed using the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \] where \( R \) is the Rydberg constant, \( n \) is an integer (3, 4, 5, ...), and \( 2^2 = 4 \). ### Step 2: Identify the Wavelengths From the formula, we can rearrange it to find \( \lambda \): \[ \lambda = \frac{1}{R \left( \frac{1}{4} - \frac{1}{n^2} \right)} \] - **Shortest Wavelength (\( \lambda_{\text{min}} \))**: This occurs when \( n \) is at its minimum value, which is 3. - **Longest Wavelength (\( \lambda_{\text{max}} \))**: This occurs when \( n \) approaches infinity. ### Step 3: Calculate the Shortest Wavelength For \( n = 3 \): \[ \lambda_{\text{min}} = \frac{1}{R \left( \frac{1}{4} - \frac{1}{3^2} \right)} = \frac{1}{R \left( \frac{1}{4} - \frac{1}{9} \right)} \] Calculating the term inside the parentheses: \[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \quad \Rightarrow \quad \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \] Thus, \[ \lambda_{\text{min}} = \frac{1}{R \cdot \frac{5}{36}} = \frac{36}{5R} \] ### Step 4: Calculate the Longest Wavelength For \( n \to \infty \): \[ \lambda_{\text{max}} = \frac{1}{R \left( \frac{1}{4} - 0 \right)} = \frac{1}{R \cdot \frac{1}{4}} = \frac{4}{R} \] ### Step 5: Find the Ratio of Largest to Shortest Wavelength Now we can find the ratio: \[ \text{Ratio} = \frac{\lambda_{\text{max}}}{\lambda_{\text{min}}} = \frac{\frac{4}{R}}{\frac{36}{5R}} = \frac{4}{R} \cdot \frac{5R}{36} = \frac{20}{36} = \frac{5}{9} \] ### Conclusion Thus, the ratio of the largest to shortest wavelength in the Balmer series of hydrogen spectra is: \[ \frac{5}{9} \]

To find the ratio of the largest to shortest wavelength in the Balmer series of hydrogen spectra, we can follow these steps: ### Step 1: Understand the Rydberg Formula The wavelength (\( \lambda \)) for the Balmer series can be expressed using the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \] where \( R \) is the Rydberg constant, \( n \) is an integer (3, 4, 5, ...), and \( 2^2 = 4 \). ...
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