JEE Main 2022 · Physics · Atoms, Nuclei and Dual Nature of Radiation · easy

A hydrogen-like ion emits a photon of energy 108.8 eV when its electron makes a transition from n=3 to n=1. Taking the hydrogen ground-state energy…

Numerical answer

A hydrogen-like ion emits a photon of energy

108.8\ \mathrm{eV}

when its electron makes a transition from

n=3

n=1

. Taking the hydrogen ground-state energy as

-13.6\ \mathrm{eV}

, the atomic number of the ion is

Show answer and solution

Answer: 3

For a hydrogen-like ion the energy levels are

E_n=-13.6\dfrac{Z^{2}}{n^{2}}\ \mathrm{eV}

. The photon energy is

\Delta E=13.6Z^{2}\left(\dfrac1{1^{2}}-\dfrac1{3^{2}}\right)=13.6Z^{2}\cdot\dfrac89=12.089Z^{2}\ \mathrm{eV}

. Setting this equal to

108.8

Z^{2}=\dfrac{108.8\times9}{13.6\times8}=9

, so

Z=3

. The ion is

\mathrm{Li^{2+}}

. Because the energy scales as

Z^{2}

, the ratio to the hydrogen value

13.6\times\frac89=12.09\ \mathrm{eV}

9

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