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In an hydrogen like atom an electron is in the excited state of quantum number n = 4. This electron can emit the maximum energetic photon of energy 204 eV. The atomic number Z of this hydrogen like atom is

In an hydrogen like atom an electron is in the excited state of quantum number n = 4. This electron can emit the maximum energetic photon of energy 204 eV. The atomic number Z of this hydrogen like atom is

Grade:12

1 Answers

Arun
25763 Points
one year ago
Please find the answer to your question
 
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/ 3– 1/ 42) = 13. 6 x 42 (7 / 9 x 16)
= 10.5 eV.
Hence n = 2, Z = 4, Emin = 10.5 ev

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