Question

The ground state energy of hydrogen atom is $-13.6eV$.
(i) What is the kinetic energy of an electron in the $2^{nd}$ excited state?

Answer 1

$\text{2nd excited state means 3rd normal state or n=3 state}$

total energy of $n^{th}level$ is $E_n=E_0/n^2$ ,

given that $E_0=-13.6eV$

So total energy for 3rd state will be $E_3=-13.6/9=-1.51eV$

We know that $\text{total energy=-kinetic energy so kinetic energy=-(-1.51eV)=1.51eV}$.