Question

The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is $3.4eV.$ Determine the de-Brogile wavelength associated with it.

Answer 1

De-Broglie wavelength is given as,

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

$\lambda =\frac{6.626\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 3.4\times 1.6\times {10}^{-19}}}$

$\lambda =0.66\times {10}^{-9}m$

Thus, de-Broglie wavelength is $0.66\times {10}^{-9}m$.