Question

If the velocity of the electron in Bohrs first orbit is $2.19\times {10}^6{ms}^{-1}$, calculate the de Broglie wavelength associated with it.

Answer 1

$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}}{9.11\times {10}^{-31}\times 2.19\times {10}^6}=3.32\times {10}^{-10}m=332pm$