The kinetic energy of an electron in the 2nd excited state in hydrogen atom, if its ground state energy is -13.6 eV, will be

1.51 eV

The correct answer is Option (3) → 1.51 eV

Energy of the n-th level in hydrogen atom:

$E_n = -\frac{13.6}{n^2} \, \text{eV}$

2nd excited state corresponds to $n = 3$:

$E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV}$

Kinetic energy of the electron:

$K = -E_n = -(-1.51) = 1.51 \, \text{eV}$

Kinetic energy = 1.51 eV