Question

The potential energy of an electron in He⁺ ion is

– 12.09 eV. Indicate in which excited state, the electron is present –

• A. First
• B. Second
• C. Third
• D. Fourth

Answer 1

Answer: (B)

Second

Answer 2

P.E. = 2T.E.

̃ – 12.09 = – 2 × 13.6 $\frac{z^{2}}{n^{2}}$

̃ n² = $\frac{2 \times 13.6 \times 2^{2}}{12.09}$= 9

̃ n = 3