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A hydrogen like atom (atomic number Z) is in higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 and 17.0 eV respectively. Alternately, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively.

A hydrogen like atom (atomic number Z) is in higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 and 17.0 eV respectively. Alternately, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV respectively.

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
7 years ago
Hello Student,
Please find the answer to your question
For hydrogen like atoms
En - 13.6 / n2 Z2 eV / atom
Given En – E2 = 10.2 + 17 = 27.2 eV …(i)
En – E3 = 4.24 + 5.95 = 10.2 eV
∴ E3 – E2 = 17
But E3 – E2 = - 13.6 / 9 Z2 – ( - 13.6 /4 Z2)
= - 13.6 Z2 [1/9 – 1/4]
= - 13.6 Z2 [4 – 9 / 36] = 13.6 x 5 / 36 z2
∴ 13.6 x 5 / 36 Z2 = 17 ⇒ Z = 3
En – E2 = 13.6 n2 x 32 – [- 13.6 / 22 x 32]
= - 13.6 [ 9 / n2 – 9 / 4] = - 13.6 x 9 [4 – n2 / 4n2] …(ii)
From eq. (i) and (ii),
- 13. 6 x 9 [ 4 – n2 / 4n2] = 27.2
⇒ - 122.4 (4 – n2) = 108 . 8 n2
⇒ n2 = 489.6 / 13.6 = 36 ⇒ n = 6
Thanks
Kevin Nash
askIITians Faculty

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