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`

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

`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=(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