A hydrogen-like atom of atomic number Z is in an excited state of quantum number 2 n . It can emit a maximum energy photon of 204 eV . If it makes a transition to the quantum state n , a photon of energy 40 . 8 eV is emitted. Calculate the atomic number Z . Ground state energy of hydrogen atom is - 13 . 6 eV

Step-by-step solution

Let ground state energy (in eV) be E 1 Then, from the given condition E 2 n - E 1 = 204 eV Or E 1 4 n 2 - E 1 = 204 eV Or E 1 1 4 n 2 - 1 = 204 eV ... (i) And E 2n - E n = 40. 8 eV Or E 1 4 n 2 - E 1 n 2 = 40.8e V or E 1 = - 3 4 n 2 = 40.8 eV .... (ii) From Eqs. (i) and (ii), 1 - 1 4 n 3 4 n 2 = 5 Or 1 = 1 4 n 2 + 1 5 4 n 2 Or 4 n 2 = 1 or n = 2 From Eq. (ii), E 1 = - 4 3 n 2 40.8 eV = - 4 3 ( 2 ) 2 ( 40.8 ) e V or E 1 = - 217 . 6 eV E 1 = - ( 13 . 6 ) Z 2 Therefore Z 2 = E 1 - 1 3 · 6 = - 2 1 7 - 6 - 1 3 · 6 = 1 6