A macroscopic sample of He ⁺ ions undergoes an electronic transition from the n=4 to the n=2 state. If the total energy released by the entire sample during this process is 8.16 J, what is the total number of photons emitted? (Given: 1 eV = 1.6 10⁻¹⁹ J)

Step-by-step solution

The energy of a single photon emitted during the transition from n=4 to n=2 for a hydrogen-like species is given by the Bohr formula: E = 13.6 Z^2 ( 1 n_f^2 - 1 n_i^2 ) eV For He ⁺ , the atomic number Z = 2 . Substituting the values: E = 13.6 (2)^2 ( 1 2^2 - 1 4^2 ) eV E = 13.6 4 ( 1 4 - 1 16 ) = 13.6 4 3 16 = 10.2 eV Convert this energy into Joules: E_p = 10.2 1.6 10⁻¹⁹ J = 16.32 10⁻¹⁹ J The total energy released by the sample is E_ total = 8.16 J . The total number of photons emitted, n = E_ total E_p n = 8.16 16