# assume that the potential energy of a hydrogen atom in its ground state is zero . then its energy in the first excited state will be

SAGAR SINGH - IIT DELHI
879 Points
12 years ago

Dear vikas,

$r_n = {n^2\hbar^2\over Zk_e e^2 m_e}$
The smallest possible value of r is called the Bohr radius and is equal to:
$r_1 = {\hbar^2 \over (k_e e^2) m_e} = 0.529 \times 10^{-10} \mathrm{m}$
The energy of the n-th level is determined by the radius:
$E = -{k_e e^2 \over 2r_n } = - { (k_e e^2)^2 m_e \over 2\hbar^2 n^2} = {-13.6 \mathrm{eV} \over n^2}$