Question

Calculate the de-Broglie wavelength associated with the electron in the $2^{nd}$ excited state of the hydrogen atom. The ground state energy of the hydrogen atom is $13.6\text{eV}$.

Answer 1

Energy of electron in ground state = $13.6\text{eV}$

For the second excited state $n=3$

The kinetic energy of an electron in a second excited state

$E=\frac{13.6}{(3{)}^2}=1.51\text{eV}$

the the the the the the the the de-Broglie wavelength of an electron in a second excited state

$\lambda =\frac{h}{\sqrt{2km}}$

$\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 1.51\times 1.6\times {10}^{-19}\times 9.1\times {10}^{-31}}}\text{m}$

$\lambda =\frac{6.63\times {10}^{-34}}{6.63\times {10}^{-25}}\text{m}$

$\lambda =1\times {10}^{-9}\text{m}$

$\lambda =10\times {10}^{-10}\text{m}$

$\lambda =10.0{\text{A}}^{\circ }$