Question

The ground state energy of hydrogen atom is - 13.6e V.
(A) What is the potential energy of an electron in the 3rd excited state?
(B) If the electron jump to the ground state from the 3rd excited state,calculate the wavelength of the photon emitted.

Answer 1

Potential energy of an electron in ${\text{n}}^{th}$ orbit
$V\left(r\right)=-\frac{Z^2{e}^{4}m}{4{ϵ}_0^2{h}^2}\frac{1}{n^2}$

$V\left(r\right)=-\frac{27.2Z^2}{n^2}eV$

For hydrogen (Z=1), third excited state (n=3)

(A) Potential Energy, $V\left(r\right)=-\frac{27.2{\left(1\right)}^2}{{\left(3\right)}^2}eV=-3.022eV$

Energy in photon = Change in Potential energy from ground state.

$\frac{hc}{\lambda }=-27.2Z^2(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

$\frac{hc}{\lambda }=-27.2(\frac{1}{3^2}-\frac{1}{1^2})=24eV$

$\lambda =\frac{hc}{24eV}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{24\times 1.6\times {10}^{-19}}=517\mu m$

(B) Hence, wavelength of photon is $517\mu m$