Most Important Selected Qs for JEE AdvancedPhysicsAtomic Physics
Consider a hydrogen like atom whose energy in n^ th excited state is given by E_n=- 13.6 n^2 Z^2 when this excited atom makes a transition from an excited state to ground state. The most energetic photons have energy E _ =52.224 eV and the least energetic photons have energy E _ =1.224 eV . Find the atomic number of atom.
Correct answer
2
Step-by-step solution
Max" energy is liberated for transition E_n E₁ and minimum energy for E_n E_ n-1 Hence, E₁ n^2 - E₁ 12 =52.224 eV and E₁ n^2 - E₁ (n-1)^2 =1.224 eV Solving we get, E₁=-54.4 eV and n=5 hence, E₁=- 13.6 Z^2 12 =-54.4Z=2