Question

The mass of an electron is 9.11×10−31 kg. If the de broglie wavelength for an electron in a hydrogen atom is 3.31×10−10 m

Answer 1

Answer: The electron moves
`0.73\%` less fast than light

Step-by-step explanation:

The complete question is written below:

The mass of an electron is
`9.11(10)^(-31) kg` . If the de Broglie wavelength for an electron in a hydrogen atom is
`3.31(10)^(-10) m`, how fast is the electron moving relative to the speed of light? The speed of light is
`3(10)^8 m/s`.

The De Broglie wavelength equation is:

`\lambda_(e)=(h)/(p_(e))` (1)

Where:

`\lambda_(e)=3.31(10)^(-10) m` is the de broglie wavelength for an electron

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

`p_(e)` is the momentum of the electron

On the other hand, the momentum of the electron is given by:

`p_(e)=m_(e)V_(e)` (2)

Where:

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

`V_(e)` is the velocity of the electron

Substituting (2) in (1):

`\lambda_(e)=(h)/(m_(e)V_(e))` (3)

Isolating
`V_(e)`:

`V_(e)=(h)/(m\lambda_(e))` (4)

`V_(e)=(6.626(10)^(-34)J.s)/((9.11(10)^(-31) kg)(3.31(10)^(-10) m))`

Finally:

`V_(e)=2,197,379.461 m/s \approx 2.20(10)^(6) m/s` This is the velocity of the electron

Calculating the ratio between the velocity of the electron and the velocity of a photon:

`((2.20(10)^(6) m/s)/(3(10)^8 m/s))100\%=(0.0073)(100)=0.73\%`

Therefore, the electron moves
`0.73\%` less fast than the photon (light).