# Chapter 6 - Electronic Structure of Atoms - Exercises: 6.50

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.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.