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

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:

$\lambda_{e}=\frac{h}{p_{e}}$ (1)

Where:  

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

$h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}$ 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):

$\lambda_{e}=\frac{h}{m_{e}V_{e}}$ (3)

Isolating $V_{e}$:

$V_{e}=\frac{h}{m\lambda_{e}}$ (4)

$V_{e}=\frac{6.626(10)^{-34}J.s}{(9.11(10)^{-31} kg)(3.31(10)^{-10} m)}$

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:

$(\frac{2.20(10)^{6} m/s}{3(10)^8 m/s})100\%=(0.0073)(100)=0.73\%$

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