Question

The ground state energy of the $H_2$ atom is $13.6eV$? what is the largest wave length of photon that can ionize a $H_2$ atom is its ground state?

Answer 1

$E_n=-\frac{13.6}{n^2}ev$.

$E_1=-13.6ev$

longest wavelength

$\therefore$ Required energy to ionize $H_2$ atom in its ground state is $\pm 13.6ev$
$\therefore E=13.6ev$ $E=hv=\frac{hc}{\lambda }$
$\lambda =\frac{hc}{E}$
$\lambda =\frac{12400}{E\left(inev\right)}$
$\therefore \lambda =\frac{12400}{13.6}=911.76\mathring{A}$
$\therefore$ longest wavelength =$921.76\mathring{A}$