In hydrogen atom, energy of first excited state is –3.4 eV. Find out KE

In hydrogen atom, energy of first excited state is –3.4 eV. Find out KE of the same orbit of Hydrogen atom

Options

(a) +3.4 eV
(b) +6.8 eV
(c) –13.6 eV
(d) +13.6 eV

Correct Answer:

+3.4 eV

Explanation:

No explanation available. Be the first to write the explanation for this question by commenting below.

admin:

View Comments (1)

Related Questions

  1. The law of equilibrium was first given by
  2. When cold potassium permanganate (KMnO₄) is added to ethylene gives
  3. Which is purified by steam distillation
  4. The number of acidic protons in H₃PO₃ are
  5. Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol.