Question

Calculate the PE, KE and the total energy ( eV ) of an electron of H 2 atom revolving
in the first Bohr orbit of radius 0.5 A.​

Answer 1

Answer:

total energy=-13.6 ev

KE= 13.6 ev

PE = -27.2 ev

Explanation:

we know  that total energy is -13.6 z²/n²

now n=1 ( given)

and Z= 1  ( hydrogen atom)

therefore total energy is -13.6 ev                                               (i)

again kinetic energy is   ( -total energy )

                      therefore    -(-13.6)                    (putting TE as -13.6 ev from i)

                               KE  =  13.6 ev    

again PE =  2(TE)

therefore PE= -27.2 ev  

nice question              

Answer 2

Answer:

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Explanation: