In $$hydrogen-like$$ atom (atomic$$ number A is in a higher excited state of quantum number n. This excited atom can make a transition to first excited state by emitting photons of energies $$10.20$$ eV and $$17.00$$ eV respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies $$4.25$$ eV and $$5.95$$ eV respectively. Determine values of n $$(ionization$$ energy of hydrogen atom $$=$$ $$13.6$$ $$eV).

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Given that En−$$E2=10.2+17.0=27.2eV$$ and En−$$E3=4.25+5.95=10.2eV$$⇒$$E3$$−$$E2=27.2$$−$$10.2=17eV$$⇒$$1.89Z2=17eV$$⇒$$Z=3$$ Now we use $$27.2=13.6Z214$$−$$1n2$$⇒$$n=6$$

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