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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 :

cryssatemp

Answer: The electron moves [tex]0.73\%[/tex] less fast than light

Explanation:

The complete question is written below:

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

The De Broglie wavelength equation is:

[tex]\lambda_{e}=\frac{h}{p_{e}}[/tex] (1)

Where:  

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

[tex]h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}[/tex] is the Planck constant

[tex]p_{e}[/tex] is the momentum of the electron

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

[tex]p_{e}=m_{e}V_{e}[/tex] (2)

Where:

[tex]m_{e}=9.11(10)^{-31} kg[/tex] is the mass of the electron

[tex]V_{e}[/tex] is the velocity of the electron

Substituting (2) in (1):

[tex]\lambda_{e}=\frac{h}{m_{e}V_{e}}[/tex] (3)

Isolating [tex]V_{e}[/tex]:

[tex]V_{e}=\frac{h}{m\lambda_{e}}[/tex] (4)

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

Finally:

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

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

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

Therefore, the electron moves [tex]0.73\%[/tex] less fast than the photon (light).

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