If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^{6} \mathrm{~ms}^{-1}$, calculate the de Broglie wavelength associated with it.
According to de Broglie's equation,
$\lambda=\frac{h}{m v}$
Where,
$\lambda=$ wavelength associated with the electron
$h=$ Planck's constant
$m=$ mass of electron
$v=$ velocity of electron
Substituting the values in the expression of $\lambda$ :
$\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(2.19 \times 10^{6} \mathrm{~ms}^{-1}\right)}$
$=3.32 \times 10^{-10} \mathrm{~m}=3.32 \times 10^{-10} \mathrm{~m} \times \frac{100}{100}$
$=332 \times 10^{-12} \mathrm{~m}$
$\lambda=332 \mathrm{pm}$
$\therefore$ Wavelength associated with the electron $=332 \mathrm{pm}$