Question

If an electron jumps from the 3rd orbit to the 2nd orbit, its wavelength is \(\lambda\). Then the wavelength of the electron when it jumps from the 4th orbit to the 3rd orbit in terms of \(\lambda\) is:

Hint:

To find the relationship between wavelengths in different transitions, compare the terms \(\frac{1}{n_1^2} - \frac{1}{n_2^2}\) using the Rydberg formula.

Answer 1

Step 1: Using the Rydberg formula for wavelength. 
The wavelength \(\lambda\) of the photon emitted during an electronic transition is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \] where: \(R\): Rydberg constant, \(n_1\): Final orbit number, \(n_2\): Initial orbit number (\(n_2>n_1\)). 

Step 2: For the transition from the 3rd orbit (\(n_2 = 3\)) to the 2nd orbit (\(n_1 = 2\)). 
The wavelength is given as \(\lambda\). Substituting into the Rydberg formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right). \] Simplify the terms: \[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36}. \] 

Step 3: For the transition from the 4th orbit (\(n_2 = 4\)) to the 3rd orbit (\(n_1 = 3\)). 
The wavelength for this transition, say \(\lambda'\), is: \[ \frac{1}{\lambda'} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right). \] Simplify the terms: \[ \frac{1}{\lambda'} = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = R \cdot \frac{7}{144}. \] 

Step 4: Relating \(\lambda'\) to \(\lambda\). 
Divide the two equations: \[ \frac{1}{\lambda'} \div \frac{1}{\lambda} = \frac{R \cdot \frac{7}{144}}{R \cdot \frac{5}{36}}. \] Simplify: \[ \frac{\lambda}{\lambda'} = \frac{\frac{7}{144}}{\frac{5}{36}} = \frac{7}{144} \cdot \frac{36}{5} = \frac{7 \cdot 36}{144 \cdot 5} = \frac{7}{20}. \] Thus: \[ \lambda' = \frac{20}{7} \lambda. \] 

Step 5: Conclusion. 
The wavelength of the electron when it jumps from the 4th orbit to the 3rd orbit is: \[ \boxed{\frac{20}{7} \lambda}. \]