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A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two 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 energy 4.25 eV and 5.95 eV respectively. Determine the value of Z. [Ionization energy of hydrogen atom = 13.6V]

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Detailed Solution

From the de-excitation to first excited state we get,

$10 .2 + 17 = 13 .6 \times Z^{2} (\frac{1}{2^{2}} - \frac{1}{n^{2}})$ ...(1)

from the de-excitation to second excited state we get,

$4 .25 + 5 .95 = 13 .6 \times Z^{2} (\frac{1}{3^{2}} - \frac{1}{n^{2}})$ ...(2)
Solving the above two equation we get, Z = 3 and n = 6.

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