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Q.

If first ionization potential of a hypothetical atom is 16 V, then the first excitation potential will be

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a

12 V

b

10.2 V

c

14 V

d

16 V

answer is B.

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Detailed Solution

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$Given, I . P_{1} = 16 eV E_{1} = - {IP}_{1} = - 16 eV First excitation is 2 \to 1 E_{n} = \frac{E_{1}}{n^{2}} = - \frac{16}{n^{2}} E_{2 \to 1} = 16 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = 16 (\frac{1}{1} - \frac{1}{4}) = 16 \times \frac{3}{4} = 12 eV$

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If first ionization potential of a hypothetical atom is 16 V, then the first excitation potential will be