Atomic StructureHard

Question

Some hydrogen-like atoms in ground state absorbs ‘n’ photons having same the energy and on de-excitement, it emits exactly ‘n’ photons. The energy of absorbed photon may be

Options

A.91.8 eV
B.40.8 eV
C.48.4 eV
D.54.4 eV

Solution

Per atom only one photon is emitted out and hence, the concerned transition is 2 → 1.

$\Delta E = 13.6Z^{2}\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) = 10.2Z^{2}\text{ eV}$

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