What is the de Broglie wavelength of an electron traveling at 2.0 * 10^8 m/s?

Jun 24, 2017

$\lambda = 3.64 \cdot {10}^{-} 12$ $\text{m}$

Explanation:

de Broglie wave equation$\to$ $\lambda = \frac{h}{p}$

...where

• $\lambda$ is the wavelength in $\text{m}$.

• $p$ ($\text{mass"(m)*"velocity} \left(v\right)$) is momentum
(electron mass = $9.109 \cdot {10}^{- 31}$ $\text{kg}$).

• $h$ is Planck's constant =6.626*10^-34 "J"("joule")*"s"("second").
(1 Joule = ${\text{1 kg"cdot"m"^2"/s}}^{2}$)

Resolving...

$\lambda = \frac{6.626 \cdot {10}^{-} 34 \text{J"cdot"s}}{m v}$

lambda=(6.626*10^-34 "J"cdot"s")/((9.109*10^-31 "kg")(2.0*10^8 "m/s"))

$\lambda = \left(6.626 \cdot {10}^{-} 34 \text{kg"cdot"m"^2"/s")/(18.2*10^-23 "kg"cdot"m/s}\right)$

Here, everything is cancelled except $\text{m}$.

$\lambda = 3.64 \cdot {10}^{-} 12$ $\text{m}$