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. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is______.

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answer is 3.

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Detailed Solution

According to question, the photon emitted $n = 2 \to n = 1$ transition has energy 74.8 eV higher than the photon emitted $n = 3 \to n = 2$

$∴ Δ E_{2 - 1} = 74.8 + Δ E_{3 - 2}$

$13.6 z^{2} [1 - \frac{1}{4}] = 74.8 + 13.6 z^{2} [\frac{1}{4} - \frac{1}{9}] ∴ z = 3$

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