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

According to de Broglie's equation, $\lambda = \frac{h}{mv}$.
Where:
$\lambda = \text{wavelength associated with the electron}$
$h = 6.626 \times 10^{-34} \, J s \text{ (Planck's constant)}$
$m = 9.109 \times 10^{-31} \, kg \text{ (mass of electron)}$
$v = 2.19 \times 10^{6} \, m s^{-1} \text{ (velocity of electron)}$
Substituting the values in the expression of $\lambda$:
$\lambda = \frac{6.626 \times 10^{-34} \, J s}{(9.109 \times 10^{-31} \, kg)(2.19 \times 10^{6} \, m s^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{1.9948 \times 10^{-24}} \, m$
$\lambda \approx 3.32 \times 10^{-10} \, m$
Since $1 \, pm = 10^{-12} \, m$, we have:
$\lambda = 332 \times 10^{-12} \, m = 332 \, pm$.
Therefore, the de Broglie wavelength associated with the electron is $332 \, pm$.