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Q.

1st excitation potential for the H-like (hypothetical sample is 24 V. Then:

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a

2nd excitation potential of the sample is 32×89V

b

binding energy of 3rd excited state is 2 eV

c

ionization energy of the sample is 36 eV

d

ionization energy of the sample is 32 eV

answer is B, C, D.

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Detailed Solution

The ionization energy;

(IE)×112−122=24

⇒IE=32eV BE 3rd excited state =3216=2 eV

2nd IP=32×112−132=32×89V

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