In hydrogen atom, energy of first excited state is-3.4 eV. Then find out KE of the same orbit of hydrogen atom.
Answers
Answered by
5
Answer:total energy E = T + V
where
T = kinetic energy and V is potential energy ,
the potential is V= - k. e^2 / r and T = (1/2)m. v^2
however the centripetal force m.v ^2 / r = k.e^2/(r^2)
therefore T =(1/2)m. v^2 = (1/2) .k. e^2 / r = - V/2
so, the magnitude of V = 2.T and is negative in character.
E = -3.4 eV = T -2.T = -T therefore the kinetic energy must be 3.4 ev
Explanation:
Similar questions