# What is the de Broglie wavelength for an electron that travels at 90% the speed of light?

Aug 29, 2017

See below

#### Explanation:

First we need to know the formula. It is, λ = h/(mv)
we know speed of light is $3.00 \times {10}^{8}$$m$/$s$, then we need to find what 90% of that is.
To find that we do this,

$0.90 c = 3.00 \times {10}^{8} \text{m/s} \times 0.90$

$= 2.7 \times {10}^{8}$ $\text{m"//"s}$

λ = h/(mv)

λ= wavelength

$h =$ Planck's constant, which is $6.63 \times {10}^{-} 34$ ${\text{m}}^{2} \cdot$$\text{kg"//"s}$

$m =$ mass of electron, which $9.11 \times {10}^{-} 31 \text{kg}$

$v =$ velocity

lambda =(6.63 xx 10^-34 "kg"cdot"m"^2//"s") / ((9.11xx10^-31 "kg") xx (2.7xx10^8 "m/s"))

$\implies \lambda = 2.70 \times {10}^{-} 12 \text{m}$