Question

Which one of the following corresponds to the wavelength of line spectrum of H atom in its Balmer series? (R = Rydberg constant)

• A. \(\frac{9}{8R}\)
• B. \(\frac{100}{21R}\)
• C. \(\frac{25}{24R}\)
• D. \(\frac{16}{15R}\)

Hint:

The Balmer series corresponds to transitions where the electron falls to the \(n=2\) level.

Answer 1

The Correct Option is B

To determine which option corresponds to the wavelength of a line in the Balmer series of the hydrogen atom, we use the Rydberg formula for the Balmer series: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \] where: - \( \lambda \) is the wavelength, - \( R \) is the Rydberg constant, - \( n \) is an integer greater than 2 (since the Balmer series corresponds to transitions to the \( n=2 \) level). Analysis of Each Option: 1. Option (1): \(\frac{9}{8R}\) - Rearrange to find \( \lambda \): \[ \lambda = \frac{9}{8R} \] - Using the Rydberg formula: \[ \frac{1}{\lambda} = \frac{8R}{9} = R \left( \frac{1}{4} - \frac{1}{n^2} \right) \] \[ \frac{8}{9} = \frac{1}{4} - \frac{1}{n^2} \] \[ \frac{1}{n^2} = \frac{1}{4} - \frac{8}{9} = \frac{9 - 32}{36} = -\frac{23}{36} \] - This is not possible since \( \frac{1}{n^2} \) cannot be negative. 2. Option (2): \(\frac{100}{21R}\) - Rearrange to find \( \lambda \): \[ \lambda = \frac{100}{21R} \] - Using the Rydberg formula: \[ \frac{1}{\lambda} = \frac{21R}{100} = R \left( \frac{1}{4} - \frac{1}{n^2} \right) \] \[ \frac{21}{100} = \frac{1}{4} - \frac{1}{n^2} \] \[ \frac{1}{n^2} = \frac{1}{4} - \frac{21}{100} = \frac{25 - 21}{100} = \frac{4}{100} = \frac{1}{25} \] \[ n^2 = 25 \implies n = 5 \] - This is a valid transition (\( n=5 \) to \( n=2 \)).Option (2) is valid. 3. Option (3): \(\frac{25}{24R}\) - Rearrange to find \( \lambda \): \[ \lambda = \frac{25}{24R} \] - Using the Rydberg formula: \[ \frac{1}{\lambda} = \frac{24R}{25} = R \left( \frac{1}{4} - \frac{1}{n^2} \right) \] \[ \frac{24}{25} = \frac{1}{4} - \frac{1}{n^2} \] \[ \frac{1}{n^2} = \frac{1}{4} - \frac{24}{25} = \frac{25 - 96}{100} = -\frac{71}{100} \] - This is not possible since \( \frac{1}{n^2} \) cannot be negative. 4. Option (4): \(\frac{16}{15R}\) - Rearrange to find \( \lambda \): \[ \lambda = \frac{16}{15R} \] - Using the Rydberg formula: \[ \frac{1}{\lambda} = \frac{15R}{16} = R \left( \frac{1}{4} - \frac{1}{n^2} \right) \] \[ \frac{15}{16} = \frac{1}{4} - \frac{1}{n^2} \] \[ \frac{1}{n^2} = \frac{1}{4} - \frac{15}{16} = \frac{4 - 15}{16} = -\frac{11}{16} \] - This is not possible since \( \frac{1}{n^2} \) cannot be negative. Final Answer: The wavelength corresponding to a line in the Balmer series of the hydrogen atom is: \[ \boxed{\frac{100}{21R}} \] This corresponds to option (2).