In hydrogen atom, energy of the first excited state is -3.4 eV. Calculate the kinetic energy of the same orbit of hydrogen atom

In hydrogen atom, energy of the first excited state is -3.4 eV. Calcul

1.	In hydrogen atom, energy of the first excited state is -3.4 eV. Calculate the kinetic energy of the same orbit of hydrogen atom
Answer» Solution :Total energy of the electron in n =2 (1st excited state) `= -3.4 eV` `E_("total") = K.E. + P.E. = (1)/(2) m v^(2) + ((-Ze^(2))/(R))` But force of altraction on the NUCLEUS = CENTRIPETAL force i.e., `(Ze^(2))/(r^(2)) = (mv^(2))/(r) or mv^(2) = (Ze^(2))/(r)` `:. E_("total") = (1)/(2) (Ze^(2))/(r) - (Ze^(2))/(r) = - (Ze^(2))/(2r) = -3.4 eV` (GIVEN) `:. K.E. = (1)/(2) mv^(2) = (1)/(2) (Ze^(2))/(r) = 3.4 eV`