JEE Challenger
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PhyMedAdv 2018 • Paper 2

Atomic Number Determination from Hydrogen Like Emission Spectrum Transitions

Consider a hydrogen-like single-electron ion with atomic number ZZ. The photon emitted in the n=2→n=1n = 2 \rightarrow n = 1 transition has an energy 74.8 eV74.8\text{ eV} greater than the photon emitted during n=3→n=2n = 3 \rightarrow n = 2 transition. Knowing that the ionization energy of a hydrogen atom is 13.6 eV13.6\text{ eV}, determine the value of ZZ.

Official Numerical Answer3

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To determine the atomic number ZZ, we use the energy formula for photon emissions in a hydrogen-like atom:

ΔE=13.6Z2(1nf2−1ni2) eV\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}

The energy difference between the n=2→1n = 2 \rightarrow 1 and n=3→2n = 3 \rightarrow 2 transitions is given as 74.8 eV74.8\text{ eV}:

ΔE2→1−ΔE3→2=13.6Z2((1−14)−(14−19))=74.8 eV\Delta E_{2 \rightarrow 1} - \Delta E_{3 \rightarrow 2} = 13.6 Z^2 \left( \left( 1 - \frac{1}{4} \right) - \left( \frac{1}{4} - \frac{1}{9} \right) \right) = 74.8\text{ eV}

Simplifying the equation gives:

13.6Z2(1118)=74.8  ⟹  Z2=9  ⟹  Z=313.6 Z^2 \left( \frac{11}{18} \right) = 74.8 \implies Z^2 = 9 \implies Z = 3

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