Question

Lamb-Rutherford experiment showed that in a hydrogen atom:

• A. \( 2S_{1/2} \) state has greater energy than \( 2P_{1/2} \) state
• B. \( 2P_{1/2} \) state has greater energy than \( 2S_{1/2} \) state
• C. \( 2S_{1/2} \) and \( 2P_{1/2} \) states have the same energy
• D. \( 2P_{1/2} \) state has lower energy than \( 2P_{3/2} \) state

Hint:

The Lamb shift was one of the key pieces of evidence leading to the development of Quantum Electrodynamics (QED).

Answer 1

The Correct Option is A

The Lamb shift, discovered by Lamb and Rutherford, demonstrated that:
\[ E(2S_{1/2})>E(2P_{1/2}) \] This shift arises due to vacuum fluctuations and quantum electrodynamic effects, causing a slight energy difference between the otherwise degenerate states.