Q50 / 54 · JEE Advanced 2018 (Paper 2) · 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?

Correct answer: 3

Solution

∆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

Hence, the correct answer is 3.

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