Question

30. In hydrogen atom energy of electron in 2nd Bohr's
orbit is -3.4 eV. The kinetic energy of electron in
this orbit is
(1) -3.4 eV
(2) +3.4 eV
(3) -6.8 eV
(4) +1.7 eV​

Answer 1

wo know kinetic energy is equal to -13.6×z^2/n^2

here Z=1 and n=2

so K.E =+13.6×1/4

K.E=+3.4ev

some important formulas

1)KE =+13.6×z^2/n^2

2)PE=-27.2×Z^/n^2

so total energy (E)=K.E+P.E

=13.6z^/n^2 - 27.2z^/n^2

=-13.6z^/n^2

hope it helps you

be brainly