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Question

A hydrogen like atom with atomic number Z is in higher excited state of quantum number 'n' . This excited state atom can make a transition to the first excited state by successively emitting two photons of energies 10 eV and 68.2eV respectively. Alternatively the atom from the same the same excited state can make a transition to the 2nd excited state by emitting two photons of energies 4.25eV and 5.95eV respectively. Calculate the value of Z

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Solution

We have, energy of the electron in the nth energy state in any atom is given by,
E=13.6z2n2eV

Let the intermediate energy state in the first case be x. So,
z24z2n2=10.2+1713.6=2(1)
Let the intermediate energy state in the 2nd case be y. So,
z29z2n2=4.25+5.9513.6=34(2)
(2)(1),536z2=54
z2=9,z=3

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