Answer: [tex]2.42(10)^{-11} 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_{e}[/tex] (2)
Where:
[tex]m_{e}=9.11(10)^{-31}kg[/tex] is the mass of the electron
[tex]v=\frac{1}{10}c[/tex] is the velocity of the electron (we are told it is 1/10 the speed of light [tex]c=3(10)^{8}\frac{m}{s}[/tex])
This means equation (2) can be written as:
[tex]p=m_{e}\frac{1}{10}c[/tex] (3)
Substituting (3) in (1):
[tex]\lambda=\frac{h}{m_{e}\frac{1}{10}c}[/tex] (4)
[tex]\lambda=10\frac{h}{m_{e} c}[/tex] (5)
Now, we only have to find [tex]\lambda[/tex]:
[tex]\lambda=10\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(3(10)^{8}\frac{m}{s})}[/tex] (6)
Finally:
[tex]\lambda=2.42(10)^{-11} m[/tex]>>> This is the de Broglie wavelength of an electron.
