Answers

2016-03-17T10:45:02+05:30
(i) The radius of the third orbit would be
= 0.529 × n²/Z °A
= 0.529 × 9/ 3 °A
= 1.587 °A

(ii) Energy of the electron = -13.6·Z²/n² eV
= -13.6(3)²/(3)²
= -13.6 eV
= 21.72 × 10^-19 J
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