Question

Calculate the three lowest energy levels of the hydrogen atom.

Answer 1

SOLUTION : $E_{n}=-\frac{2 \pi^{2} c^{4} m}{n^{2} h^{2}}=-\frac{2 \cdot 179 \times 10^{-11}}{n^{2}} erg / atom$

$n=1, E=\frac{-2 \cdot 179 \times 10^{-11}}{1^{2}} erg /$ atom $=-2 \cdot 179 \times 10^{-11} erg / atom$

$n=2, E=\frac{-2 \cdot 179 \times 10^{-11}}{2^{2}} erg / atom =-0 \cdot 5448 \times 10^{-11} erg / atom$

$n=3, E=\frac{-2 \cdot 179 \times 10^{-11}}{3^{2}} erg / atom =-0 \cdot 2421 \times 10^{-11} erg / atom .$