A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 and 17.0 eV respectively. Alternately, 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 the value of ‘n’. (Ionization energy of H-atom = 13.6eV)

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answer is 6.

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Detailed Solution

$Z^{2} (\frac{1}{4} - \frac{1}{n^{2}}) = 2; Z^{2} (\frac{1}{9} - \frac{1}{n^{2}}) = \frac{3}{4}$

$(10.20 + 17.00) eV = 13.6 Z^{2} \times (\frac{1}{2^{2}} - \frac{1}{n^{2}}) . . . (1)$

$(4.24 + 5.95) eV = 13.6 Z^{2} \times (\frac{1}{3^{2}} - \frac{1}{n^{2}}) . . . (2)$

On dividing equation 1 and 2,

$\begin{aligned} 1.18 = \frac{n^{2} - 4}{n^{2} - 9} \\ 0.18 n^{2} = 6.6 \\ n^{2} = 36 \\ n = 6 \end{aligned}$

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