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 solubility of BaSO₄, in water, is 2.33 ˣ 10⁻³ g/L.Its solubility product
  2. The de-Broglie wavelength associated with a ball of mass 1kg having kinetic energy
  3. If avagadro number is changed what changes
  4. Which of the following ions can cause coagulation of proteins
  5. A corked flask containing boiling water and its vapour is allowed