# What is the wavelength of an electron with a velocity of 3.85xx10^6" m/s"?

Nov 4, 2014

The wavelength of an electron with a velocity of $3.85 \times {10}^{6}$ $\text{m/s}$ is $1.89 \times {10}^{- 10}$ $\text{m}$.

#### Explanation:

The de Broglie equation relating the wavelength of an electron to its velocity is:

$\lambda$ = $\frac{h}{p}$

where

$\lambda = \text{wavelength in m}$

$h =$ Planck's constant

$p = \text{momentum} = m v$,

where:

$m = \text{kg}$, and $v = \text{m/s}$

Known/Given:

Planck's constant, $6.626 \times {10}^{- 34} \text{J"*"s}$

velocity of electron, $3.85 \times {10}^{6} \text{m/s}$

mass of electron, $9.109 \times {10}^{- 31} \text{kg}$

Unknown:

momentum $\left(p\right)$

wavelength $\left(\lambda\right)$

Equations:

$p$ = $m v$

$\lambda$ = $\frac{h}{p}$

Solution:

1. Calculate momentum.

p = (9.109xx10^(-31)"kg")xx(3.85xx10^6"m/s")= 3.507xx10^(-24)"kg"*"m/s"

2. Calculate wavelength:

$\lambda = \left(6.626 \times {10}^{- 34} \text{J"*"s")/(3.507xx10^(-24)"kg"*"m/s}\right) = 1.89 \times {10}^{- 10}$ $\text{m}$. (rounded to three significant figures)

Because "1 Joule"= "1 kg"*"m"^2/("s"^2), kilograms and seconds cancel, leaving meters.

The wavelength of an electron with a velocity of $3.85 \times {10}^{6}$ $\text{m/s}$ is $1.89 \times {10}^{- 10}$ $\text{m}$.