In the above problem the second excitation potential of the atom will be

108.8 V

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8.8 V

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208.8 V

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308.8 V

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Solution

The correct option is A 108.8 V
Second excitation potential is:
$E_{3} - E_{1} = I E - \frac{I E}{9} = \frac{8}{9} I . E = \frac{8 \times 122.4}{9} = 108.8 V$

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Similar questions

THe second excitation potential of a given atom is 12.08V. The ionisation potential must be

A. 12.08V

B. 13.6V

n = 3

20 V

122.4 V

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