Answer: [tex]1.212(10)^{-10} m[/tex]
Explanation:
The de Broglie wavelength [tex]\lambda[/tex] is given by the following formula:
[tex]\lambda=\frac{h}{p}[/tex] (1)
Where:
[tex]h=6.626(10)^{-34}\frac{m^{2}kg}{s}[/tex] is the Planck constant
[tex]p[/tex] is the momentum of the atom, which is given by:
[tex]p=m_{e}v[/tex] (2)
Where:
[tex]m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg[/tex] is the mass of the electron
[tex]v=6(10)^{6}m/s[/tex] is the velocity of the electron
This means equation (2) can be written as:
[tex]p=(9.11(10)^{-31}kg)(6(10)^{6}m/s)[/tex] (3)
Substituting (3) in (1):
[tex]\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)}[/tex] (4)
Now, we only have to find [tex]\lambda[/tex]:
[tex]\lambda=1.2122(10)^{-10} m[/tex]>>> This is the de Broglie wavelength of the electron
