# A hydrogen- like atom of atom number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) for this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is – 13. 6 eV.

Navjyot Kalra
10 years ago
Hello Student,
Energy for an orbit of hydrogen like atoms is
En = - 13. 6 Z2 / n2
For transition from 2n orbit to 1 orbit
Maximum energy = 13. 6 Z2 (1/ 1 – 1 / (2n)2)
⇒ 204 = 13. 6 Z2 (1/1 - 1/ 4n2) ….(i)
Also for transition 2n ⟶ n.
40.8 = 13. 6 Z2 (1/ n2 – 1/ 4n2) ⇒ 40.8 = 13. 6 Z2 (3 / 4n2)
⇒ 40.8 = 40.8 Z2 / 4n2 = Z2 or 2n = Z ….. (ii)
From (i) and (ii)
204 = 13.6 Z2 (1 – 1/ Z2) = 13.6 Z2 – 13.6
13. 6 Z2 = 204 + 13.6 = 217 .6
Z2 = 217.6 / 13.6 = 16, Z = 4, n = Z / 2 = 4 / 2 = 2
orbit no. = 2n = 4
For minimum energy = Transition from 4 to 3.
E = 13. 6 x 42 (1/ 32 – 1/ 42) = 13. 6 x 42 (7 / 9 x 16)
= 10.5 eV.
Hence n = 2, Z = 4, Emin = 10.5 eV
Thanks
Navjot Kalra