Question

The difference between the frequencies of second and first Paschen lines of hydrogen atom is
(\( R \) - Rydberg constant and \( c \) - speed of light in vacuum)

• A. \( \dfrac{9Rc}{16} \)
• B. \( \dfrac{16Rc}{25} \)
• C. \( \dfrac{9Rc}{400} \)
• D. \( \dfrac{3Rc}{200} \)

Hint:

Paschen series corresponds to transitions ending at \( n=3 \). Use the frequency formula \( \nu = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) to calculate spectral line differences.

Answer 1

The Correct Option is C

Step 1: Use the formula for frequency of transition in hydrogen atom: \[ \nu = Rc \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) \] Step 2: Paschen series has transitions ending at \( n_1 = 3 \).
- First Paschen line: \( n_2 = 4 \rightarrow n_1 = 3 \)
- Second Paschen line: \( n_2 = 5 \rightarrow n_1 = 3 \) 
Step 3: Calculate the frequencies. \[ \nu_1 = Rc \left( \dfrac{1}{3^2} - \dfrac{1}{4^2} \right) = Rc \left( \dfrac{1}{9} - \dfrac{1}{16} \right) = Rc \left( \dfrac{16 - 9}{144} \right) = \dfrac{7Rc}{144} \] \[ \nu_2 = Rc \left( \dfrac{1}{3^2} - \dfrac{1}{5^2} \right) = Rc \left( \dfrac{1}{9} - \dfrac{1}{25} \right) = Rc \left( \dfrac{25 - 9}{225} \right) = \dfrac{16Rc}{225} \] Step 4: Frequency difference: \[ \Delta \nu = \nu_1 - \nu_2 = \dfrac{7Rc}{144} - \dfrac{16Rc}{225} \] Find LCM of 144 and 225 = 3600, so: \[ \Delta \nu = \dfrac{175Rc}{3600} - \dfrac{256Rc}{3600} = \dfrac{-81Rc}{3600} \] We take magnitude (positive value): \[ |\Delta \nu| = \dfrac{81Rc}{3600} = \dfrac{9Rc}{400} \] Step 5: Select the correct option.
The frequency difference is \( \dfrac{9Rc}{400} \), which matches option (3).