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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answer is 6.
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Detailed Solution
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