Question

Calculate the de-Broglie wavelength of the electron orbiting in the $n=2$ states of hydrogen atom.

Answer 1

$n=2$

$K.E$ energy of the second state is:

$\frac{13.6}{n^2}=\frac{13.6}{4}$

$3.4\times 1.6\times {10}^{-19}j$

de brogile wavelength

$\lambda =\frac{h}{\sqrt{2m{E}_k}}$

$\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 3.4\times 1.6\times {10}^{-19}}}$

$=0.67nm$