Question

The energy of an electron in the ground state of hydrogen atom is −13.6 eV. The kinetic and potential energy of the electron in the first excited state will be ______.

Options

• −13.6 eV, 27.2 eV

• −6.8 eV, 13.6 eV

• 3.4 eV, −6.8 eV

• 6.8 eV, −3.4 eV

Answer 1

The energy of an electron in the ground state of hydrogen atom is −13.6 eV. The kinetic and potential energy of the electron in the first excited state will be 3.4 eV, −6.8 eV.

Explanation:

For first excited state, n = 2

`E_1 = (-13.6)/n^2`

= `(-13.6)/(2)^2`

= `(-13.6)/4`

E₁ = −3.4 eV

Now, K = −E

= − (−3.4)

= 3.4

U = 2E

= 2(−3.4)

= −6.8 eV