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Consider a hydrogen like atom whose energy in $n^{t h}$ excited state is given by $E_{n} = - \frac{13.6 Z^{2}}{n^{2}}$ . When this excited atom, makes a transition from excited state to ground state, most energetic photons have energy $E_{max} = 52.224 eV$ and least energetic photons have energy $E_{m i n} = 1 .224 eV$ . The atomic number of atom is

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Detailed Solution

Maximum energy is liberated for transition $E_{n} \to 1$ and minimum energy for $E_{n} \to E_{n - 1}$

$Hence \frac{E_{1}}{n^{2}} - E_{1} = 52.224 eV . . . . . . . . . (i) and \frac{E_{1}}{n^{2}} - \frac{E_{1}}{(n - 1)^{2}} = 1.224 eV . . . . . . . . . . (ii)$

Solving equations (i) and (ii) we get

$E_{1} = - 54.4 eV and n = 5 Now E_{1} = - \frac{13.6 Z^{2}}{1^{2}} = - 54.4 eV . Hence Z = 2$

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