- The kinetic energy of an electron in the first Bohr orbit of the hydrogen atom is 4.136 eV. Its kinetic energy in the second Bohr orbit will be:
(A) 2.068 eV (B) 8.272 eV (C) 16.544 eV (D) 33.088 eV
[JEE Mains 2018]
The answer is (B)
The kinetic energy of an electron in the nth Bohr orbit of the hydrogen atom is given by:
K = -13.6/n^2 eV
For n = 1, K = -13.6/1^2 = -13.6 eV
For n = 2, K = -13.6/2^2 = -3.4 eV
So, the kinetic energy of an electron in the second Bohr orbit will be 8.272 eV.
- The ionization potential of hydrogen atom is 13.6 eV. The energy required to remove an electron from the n = 2 state of the hydrogen atom is:
