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the binding e...

Question

the binding energy of h atom is 13.6 ev .what energy is needed for escaping the electron from first excited state of Li++ .

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Solution

${Rydberg formula for hydrogen like atom is (∵Li}^{+ +} is a Hydrogen like atom) E = R Z^{2} (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}) Now for ionisation formula becomes E = R Z^{2} (∵ n_{1} = 1 and n_{2} = \infty) Now for Hydrogen Z = 1 so E = R = 13.6 F o r L i^{+ +} Z = 3 Hence E = 13.6 \times 3^{2} = 122.4 e v$

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