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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