The total energy of an electron in the first excited state of hydrogen atom

The total energy of an electron in the first excited state of hydrogen atom is about -3.4 eV. Its kinetic energy in this state is

Options

(a) 3.4 eV
(b) 6.8 eV
(c) -3.4 eV
(d) -6.8 eV

Correct Answer:

3.4 eV

Explanation:

KE. = |(1/2) P.E.|
But P.E. is negavite
.·. Total energy = |(1/2) P.E.| – P.E. = – P.E. / 2 = – 3.4 eV
.·. K.E. = + 3.4 eV

admin:

Related Questions

  1. For the radioactive nuclei that undergo either α or β decay, which one
  2. A galvanometer having internal resistance 10Ω requires 0.01 A for a full scale
  3. A body of mass m slides down an incline and reaches the bottom
  4. The de-Broglie wavelength of an electron in the first Bohr orbit is
  5. The coefficient of mutual inductance of two coils is 6 mH. If the current flowing