Question

If Ionixation potential of an atom is 20 v then what is energy of atom in First Excited state​

Answer 1

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⟹ Excitation energy =15 V.

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Answer 2

Answer:

Hi

Explanation:

E = $13.6^{2}$

$Z^{2} = \frac{E}{13.6}\\ \\Z^{2} = \frac{20}{13.6}$

Excitation energy is given as = -13.6 *[ $\frac{Z^{2} }{4} - \frac{Z^{2} }{1}$]  ( Equation 1)

Putting $Z^{2}$ value in equation 1, We get;

Excitation energy = -13.6 * $\frac{20/4 - 20/1}{13.6}$

Excitation energy = 15 V

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