A hydrogen- like atom of atom number Z is in an ex

Question

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 · Accepted Answer

Hello Student,

Please find the answer to your question

Energy for an orbit of hydrogen like atoms is

E_n = - 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, E_min = 10.5 eV

Thanks
Navjot Kalra
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