Question

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

• A. 4 : 1
• B. 1 : 2
• C. 1 : 4
• D. 2 : 1

Answer 1

The Correct Option is A

The wavelength of a spectral line in the hydrogen spectrum is given by:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \]

where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

For the shortest wavelength in the Balmer series:

\[ n_1 = 2, \quad n_2 = \infty \]

\[ \frac{1}{\lambda_B} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = RZ^2 \left( \frac{1}{4} \right). \]

For the shortest wavelength in the Lyman series:

\[ n_1 = 1, \quad n_2 = \infty \]

\[ \frac{1}{\lambda_L} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = RZ^2 (1). \]

Taking the ratio of wavelengths:

\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{1}{RZ^2 \frac{1}{4}}}{\frac{1}{RZ^2 (1)}} = 4 : 1. \]

Thus, the ratio is:

\[ \lambda_B : \lambda_L = 4 : 1. \]

Final Answer: 4:1 (Option 1)

Answer 2

The problem asks for the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for a hydrogen atom.

Concept Used:

1. Rydberg's Formula: The wavelength \( \lambda \) of the radiation emitted when an electron in a hydrogen-like atom makes a transition from a higher energy level \( n_i \) to a lower energy level \( n_f \) is given by Rydberg's formula:

\[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

where \( R \) is the Rydberg constant and \( Z \) is the atomic number. For a hydrogen atom, \( Z = 1 \).

2. Lyman and Balmer Series:

• For the Lyman series, the electron transitions to the ground state, so the final energy level is \( n_f = 1 \). The initial energy levels are \( n_i = 2, 3, 4, \dots \).
• For the Balmer series, the electron transitions to the first excited state, so the final energy level is \( n_f = 2 \). The initial energy levels are \( n_i = 3, 4, 5, \dots \).

3. Shortest Wavelength (Series Limit): The energy of the emitted photon is \( E = \frac{hc}{\lambda} \). To have the shortest wavelength (\( \lambda_{\text{min}} \)), the energy of the transition must be maximum. This occurs when the electron transitions from the highest possible energy level, which is \( n_i = \infty \).

Step-by-Step Solution:

Step 1: Determine the shortest wavelength of the Lyman series (\( \lambda_{\text{Lyman, min}} \)).

For the Lyman series, the final state is \( n_f = 1 \). For the shortest wavelength, the initial state is \( n_i = \infty \). For hydrogen, \( Z=1 \). Substituting these values into Rydberg's formula:

\[ \frac{1}{\lambda_{\text{Lyman, min}}} = R (1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]

Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:

\[ \frac{1}{\lambda_{\text{Lyman, min}}} = R (1 - 0) = R \]

Therefore, the shortest wavelength for the Lyman series is:

\[ \lambda_{\text{Lyman, min}} = \frac{1}{R} \]

Step 2: Determine the shortest wavelength of the Balmer series (\( \lambda_{\text{Balmer, min}} \)).

For the Balmer series, the final state is \( n_f = 2 \). For the shortest wavelength, the initial state is \( n_i = \infty \). Again, for hydrogen, \( Z=1 \). Substituting these values into Rydberg's formula:

\[ \frac{1}{\lambda_{\text{Balmer, min}}} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]

This simplifies to:

\[ \frac{1}{\lambda_{\text{Balmer, min}}} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]

Therefore, the shortest wavelength for the Balmer series is:

\[ \lambda_{\text{Balmer, min}} = \frac{4}{R} \]

Final Computation & Result:

Step 3: Calculate the ratio of the shortest wavelength of the Balmer series to that of the Lyman series.

\[ \frac{\lambda_{\text{Balmer, min}}}{\lambda_{\text{Lyman, min}}} = \frac{4/R}{1/R} \] \[ \frac{\lambda_{\text{Balmer, min}}}{\lambda_{\text{Lyman, min}}} = 4 \]

Thus, the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is 4:1.