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

As per de Broglie’s equation,

$\lambda =\frac{h}{m\nu}$

Where, λ is the wavelength of the electron

h is Planck’s constant

m is the mass of the electron

v denotes the velocity of electron

Substituting these values in the expression for λ:

$\lambda =\frac{h}{m\nu }$=$\frac{(6.626\times 10^{-34})}{9.103939\times 10^{-31}kg(2.19\times 10^{6}ms^{-1})}$=$3.32\times 10^{-10}m\lambda =332pm$