Question

The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-28 g.

Answer 1

Answer: $1.212(10)^{-10} m$

Explanation:

The de Broglie wavelength $\lambda$ is given by the following formula:

$\lambda=\frac{h}{p}$ (1)

Where:

$h=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

$p$ is the momentum of the atom, which is given by:

$p=m_{e}v$ (2)

Where:

$m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg$ is the mass of the electron

$v=6(10)^{6}m/s$ is the velocity of the electron

This means equation (2) can be written as:

$p=(9.11(10)^{-31}kg)(6(10)^{6}m/s)$ (3)

Substituting (3) in (1):

$\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)}$ (4)

Now, we only have to find $\lambda$:

$\lambda=1.2122(10)^{-10} m$>>> This is the de Broglie wavelength of the electron