If '\(\lambda_1\)' and '\(\lambda_2\)' are the wavelengths of the first member of the Balmer and Paschen series, in hydrogen atom respectively, then the ratio of respective frequencies, \(f_1/f_2\) , is
Step 1: Understanding the Concept:
The spectral lines of a hydrogen atom are grouped into series based on the lower energy level involved in the transition.
Frequency and wavelength are inversely related: \( f = c/\lambda \). Step 2: Key Formula or Approach:
The Rydberg formula gives the reciprocal of wavelength: \( \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
Since frequency \( f = \frac{c}{\lambda} \), we have \( f = cR \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \). Step 3: Detailed Explanation: For the Balmer series:
The final state is \( n_f = 2 \).
The first member corresponds to a transition from the nearest upper state, so initial state \( n_i = 3 \).
Let its frequency be \( f_1 \):
\[ f_1 = cR \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = cR \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ f_1 = cR \left( \frac{9 - 4}{36} \right) = cR \left( \frac{5}{36} \right) \]
For the Paschen series:
The final state is \( n_f = 3 \).
The first member corresponds to a transition from the nearest upper state, so initial state \( n_i = 4 \).
Let its frequency be \( f_2 \):
\[ f_2 = cR \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = cR \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ f_2 = cR \left( \frac{16 - 9}{144} \right) = cR \left( \frac{7}{144} \right) \]
Finding the ratio:
Now, calculate the ratio \( \frac{f_1}{f_2} \):
\[ \frac{f_1}{f_2} = \frac{cR (5 / 36)}{cR (7 / 144)} \]
Cancel the common terms \( cR \):
\[ \frac{f_1}{f_2} = \frac{5 / 36}{7 / 144} \]
\[ \frac{f_1}{f_2} = \frac{5}{36} \times \frac{144}{7} \]
Since \( 144 = 36 \times 4 \):
\[ \frac{f_1}{f_2} = 5 \times \frac{4}{7} \]
\[ \frac{f_1}{f_2} = \frac{20}{7} \]
Step 4: Final Answer:
The ratio of respective frequencies is \( 20 : 7 \).