Answer
The characteristic wavelength of the electron is $\lambda=0.077nm$.
This wavelength is quite comparable to the size of atoms.
Work Step by Step
*The de Broglie relationship: $$\lambda=\frac{h}{mv}$$
$\lambda$: wavelength of object
$h$: Planck's constant ($h=6.626\times10^{-34} J.s$)
$m$: mass of object
$v$: velocity of object
1) Find the known variables
- Velocity of the neutron: $v=9.47\times10^6m/s$
- Mass of the electron: $m\approx9.109\times10^{-31}kg$
2) Calculate the characteristic wavelength of the electron
$\lambda=\frac{h}{mv}=\frac{6.626\times10^{-34}}{(9.109\times10^{-31})\times(9.47\times10^{6})}\approx7.681\times10^{-11}m\approx0.077nm$
3) Compare to the size of atoms
- The diameter of atoms range from $0.1$ to $0.5nm$
Therefore, the characteristic wavelength of this electron is quite comparable to the size of atoms.
