Question

kinetic and potential energy ( in eV) of electrons present in second bohr orbit of hydrogen atoms are respectively

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Answer 1

Explanation:

Hey MATE...........

We have,

|K.E|= (1/2 )|P.E|--------(i)

E= (-13.6)× (z^2/n^2) eV/atom

For n= 2, |K.E|= (-13.6)×1/4 = -3.4 eV/atom

Now,

|P.E|=2×|K.E|= 2×(-3.4) = -6.8 eV/atom

Hence, the kinetic energy and potential energy of the electron present in the second orbit of Bohr's H-atom is -3.4 eV/atom and -6.8 eV/atom respectively.

HOPE IT HELPS YOU

Answer 2

Answer: WE KNOW THAT KE = 1/2 PE AND PE = - 2KE

Explanation:

E= -13.6 ×Z²/n² eV /atom

FOE n=2 , -13.6×1²/2² eV/ atom

                = -13.6×1/4 eV/ atom = - 3.4 eV /atom

PE = 2×KE = 2×(-3.4) = - 6.8 eV /atom