Question

Energy levels of H atom are given by $E_n=-\frac{13.6}{n^2}$eV, where n is principal quantum number. Calculate the wavelength of electromagnetic radiation emitted by hydrogen atom resulting from the transition: $n=2$ to $n=1$.

Answer 1

$E_n=\frac{-13.6}{n^2}$eV.
$n=2$ to $n=1$
$\lambda =?$
$\therefore E_2=\frac{-13.6}{2^2}$
$=-3.4$eV
and $E_1=\frac{-13.6}{1^2}=-13.6$eV.
$\Delta E=E_2-E_1$
$=-3.4-(-13.6)=10.2$eV.
We know that $\Delta =\frac{hc}{\lambda }$
$\therefore \lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{10.2\times 1.6\times {10}^{-19}}$
$=1.213\times {10}^{-7}$
$=1213\stackrel{o}{A}$.