Question

is potential energy of an electron in the first orbit of Bohr's hydrogen atom is zero the total energy of the electron in the second orbit is​

Answer 1

TE in first excited state = -13.6/4 = -3.4 eV

PE in the first excited state = -6.8 eV (-PE = 2 X -TE)

For this to be zero, We must add 6.8 eV. So, add this 6.8 eV to all energy values to get the new values because shifting of zero or reference simply shifts all energy by the same value.

KE in ground state = 13.6 eV

So, in new system = 13.6 + 6.8 = 20.4 eV

Answer 2

Answer:

there is no electrons in 2nd orbit of hydrogen