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.

Question

asked Jan 21, 2020 157k views

1 Answer

Collin M · Answer 1 · 2020-01-27T15:58:22+0000

Answer:
1.212(10)^(-10) m

Step-by-step explanation:

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

$\lambda=(h)/(p)$ (1)

Where:

h=6.626(10)^(-34)(m^(2)kg)/(s) is the Planck constant

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):

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

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

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