Question

Q240) In hydrogen atom, energy of first excited state is –3.4 eV. Then find out KE of same orbit of hydrogen atom

Q240) In hydrogen atom, energy of first excited state is –3.4 eV. Then find out KE of same orbit of hydrogen atom

SOLUTION

K= −TotalEnergy(E)=3.4eV

Negative kinetic energy equals half the potential energy (−K = ½U).

Potential energy equals twice the total energy (U = 2E).

Total energy equals negative kinetic energy (E = −K).

Twice the kinetic energy plus the potential energy equals zero (2K + U = 0).

This is a key relationship for a larger problem in orbital mechanics known as the virial theorem.

Answer:(a)