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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 t o n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 t o 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

Step-by-step solution

∆ E 2 → 1 = 13.6 × Z 2 1 - 1 4 = 13.6 × Z 2 3 4 ∆ E 3 → 2 = 13.6 × Z 2 1 4 - 1 9 = 13.6 × Z 2 5 36 ∆ E 2 → 1 = ∆ E 3 - 2 + 74.8 13.6 × Z 2 3 4 = 13.6 × Z 2 5 36 + 74.8 13.6 × Z 2 3 4 - 5 36 = 74.8 Z 2 = 9 Z = + 3

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