Question

The energy of an electron in the fourth excited state of hydrogen atom is

Answer 1

Energy of nth state of hydrogen atom fourth excited state n = 5

E_n = \(\frac{-13.6}{n^2}\)

= \(\frac{-13.6}{5^2}\)

= \(\frac{-13.6}{25}\)

E₅ = -0.544 ev