Question

What is the ratio of the shortest wavelength present in the Lyman series of hydrogen spectral emissions to the shortest wavelength present in the Balmer series?

• A. \( 1 : 4 \)
• B. \( 4 : 1 \)
• C. \( 1 : 2 \)
• D. \( 2 : 1 \)

Hint:

The shortest wavelength limit for any hydrogen series can be quickly found using the shortcut formula \(\lambda_{\text{min}} = \frac{n_1^2}{R_\infty}\). This shows that the minimum wavelength scales directly with the square of the destination orbit number (\(n_1^2\)).

Answer 1

The Correct Option is A

Concept: The atomic emission spectrum of Hydrogen is split into distinct spectral series. These series describe the light emitted when an excited electron drops from a higher outer energy orbit (\(n_2\)) down to a lower inner destination orbit (\(n_1\)). The wavelengths of these emitted photons are predicted by the Rydberg formula: \[ \frac{1}{\lambda} = R_\infty \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \(\lambda\) is the emitted wavelength and \(R_\infty\) is the Rydberg constant (\(\approx 1.097 \times 10^7\,\text{m}^{-1}\)). The energy of the emitted photon is linked to its wavelength by the equation \(E = \frac{hc}{\lambda}\). This inverse relationship means that the shortest wavelength in any series corresponds to the transition with the maximum possible energy change. The largest energy jump occurs when an electron drops from the farthest possible boundary state—the ionization threshold at infinity (\(n_2 = \infty\))—down to the base series level (\(n_1\)).

Step 1: Calculate the shortest wavelength limit for the Lyman series. The Lyman series is defined by transitions where electrons drop down to the lowest ground state energy level (\(n_1 = 1\)). To find its shortest wavelength limit (\(\lambda_L\)), set the starting orbit to infinity (\(n_2 = \infty\)): \[ \frac{1}{\lambda_L} = R_\infty \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \] Since \(\frac{1}{\infty} = 0\), the equation simplifies to: \[ \frac{1}{\lambda_L} = R_\infty (1 - 0) = R_\infty \quad \Rightarrow \quad \lambda_L = \frac{1}{R_\infty} \]

Step 2: Calculate the shortest wavelength limit for the Balmer series. The Balmer series covers transitions where electrons drop down to the second energy level (\(n_1 = 2\)). Setting the starting level to infinity (\(n_2 = \infty\)) gives its shortest wavelength limit (\(\lambda_B\)): \[ \frac{1}{\lambda_B} = R_\infty \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \] \[ \frac{1}{\lambda_B} = R_\infty \left( \frac{1}{4} - 0 \right) = \frac{R_\infty}{4} \quad \Rightarrow \quad \lambda_B = \frac{4}{R_\infty} \]

Step 3: Calculate the ratio of the two shortest wavelengths. Now, find the ratio of the shortest Lyman wavelength to the shortest Balmer wavelength: \[ \frac{\lambda_L}{\lambda_B} = \frac{\frac{1}{R_\infty}}{\frac{4}{R_\infty}} \] Cancel the common Rydberg constant (\(R_\infty\)) from both the numerator and denominator: \[ \frac{\lambda_L}{\lambda_B} = \frac{1}{4} \] Therefore, the ratio of their shortest wavelengths is exactly \(1 : 4\).