Question

The ratio of wavelengths of first line (\( n_2 = 3 \)) and second line (\( n_2 = 4 \)) of Balmer series of hydrogen spectrum is

• A. 9 : 5
• B. 27 : 20
• C. 20 : 27
• D. 5 : 9

Hint:

When dealing with spectral series in the hydrogen atom, always identify the correct $n_1$ value for the specific series (e.g., $n_1=1$ for Lyman, $n_1=2$ for Balmer, $n_1=3$ for Paschen). Remember that $n_2$ is always greater than $n_1$. Simplify fractions at each step to avoid large numbers and reduce calculation errors.

Answer 1

The Correct Option is B

Step 1: Identify the Formula for Wavelength in Hydrogen Spectrum (Rydberg Formula)
For the hydrogen spectrum, the wavelength ($\lambda$) of a spectral line is given by the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where: \begin{itemize} \item $R_H$ is the Rydberg constant. \item $n_1$ is the principal quantum number of the lower energy level. \item $n_2$ is the principal quantum number of the higher energy level, where $n_2>n_1$. \end{itemize} Step 2: Define Balmer Series Parameters
For the Balmer series, the electron transitions end in the second energy level, meaning $n_1 = 2$. Step 3: Calculate Wavelength for the First Line of Balmer Series
The first line of the Balmer series corresponds to a transition from $n_2 = 3$ to $n_1 = 2$. Let this wavelength be $\lambda_1$. \[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \] Find a common denominator (36): \[ \frac{1}{\lambda_1} = R_H \left( \frac{9 - 4}{36} \right) \] \[ \frac{1}{\lambda_1} = R_H \left( \frac{5}{36} \right) \quad \cdots (1) \] Step 4: Calculate Wavelength for the Second Line of Balmer Series
The second line of the Balmer series corresponds to a transition from $n_2 = 4$ to $n_1 = 2$. Let this wavelength be $\lambda_2$. \[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] \[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{4} - \frac{1}{16} \right) \] Find a common denominator (16): \[ \frac{1}{\lambda_2} = R_H \left( \frac{4 - 1}{16} \right) \] \[ \frac{1}{\lambda_2} = R_H \left( \frac{3}{16} \right) \quad \cdots (2) \] Step 5: Determine the Ratio of Wavelengths \( \frac{\lambda_1}{\lambda_2} \)
From equation (1), $\lambda_1 = \frac{36}{5 R_H}$.
From equation (2), $\lambda_2 = \frac{16}{3 R_H}$.
Now, calculate the ratio $\frac{\lambda_1}{\lambda_2}$: \[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{36}{5 R_H}}{\frac{16}{3 R_H}} \] Cancel out $R_H$: \[ \frac{\lambda_1}{\lambda_2} = \frac{36}{5} \times \frac{3}{16} \] Simplify the fractions: \[ \frac{\lambda_1}{\lambda_2} = \frac{9 \times 4}{5} \times \frac{3}{4 \times 4} \] \[ \frac{\lambda_1}{\lambda_2} = \frac{9}{5} \times \frac{3}{4} \] \[ \frac{\lambda_1}{\lambda_2} = \frac{27}{20} \] The ratio is $27:20$. Step 6: Analyze Options
\begin{itemize} \item Option (1): 9 : 5. Incorrect. \item Option (2): 27 : 20. Correct, as it matches our calculated ratio. \item Option (3): 20 : 27. Incorrect. \item Option (4): 5 : 9. Incorrect. \end{itemize}