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JEE Advanced · Physics · 27. Atomic Physics

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n=2 to n= 1 transition has energy 74.8 eV higher than the photon emitted in the n=3 to n=2 transition. Given that the ionization energy of the hydrogen atom is 13.6 eV, what is the value of Z?

  1. A 3
  2. B 5
  3. C 7
  4. D 9
Verified Solution

Answer & Solution

Correct Answer

(A) 3

Step-by-step Solution

Detailed explanation

E2 1=13.6×Z21-14=13.6×Z234
E3 2=13.6×Z214-19=13.6×Z2536
E2 1=E3-2+74.8
13.6×Z234=13.6×Z2536+74.8
13.6×Z234-536=74.8
Z2=9
Z=+3
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