Question

What is the de Broglie wavelength of an electron that strikes the back of the face of a TV screen at 1/10 the speed of light? _____m

Answer 1

Answer:
`2.42(10)^(-11) 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_(e)` (2)

Where:

`m_(e)=9.11(10)^(-31)kg` is the mass of the electron

`v=(1)/(10)c` is the velocity of the electron (we are told it is 1/10 the speed of light
`c=3(10)^(8)(m)/(s)`)

This means equation (2) can be written as:

`p=m_(e)(1)/(10)c` (3)

Substituting (3) in (1):

`\lambda=(h)/(m_(e)(1)/(10)c)` (4)

`\lambda=10(h)/(m_(e) c)` (5)

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

`\lambda=10(6.626(10)^(-34)(m^(2)kg)/(s))/((9.11(10)^(-31)kg)(3(10)^(8)(m)/(s)))` (6)

Finally:

`\lambda=2.42(10)^(-11) m`>>> This is the de Broglie wavelength of an electron.