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

To calculate the de Broglie wavelength for a particle, or for a tennis ball for that matter, just use the equation $p = h/(lamda)$, where $p$ - the momentum...

The de Broglie wavelength ($λ$) is given by $color(blue)(|bar(ul(color(white)(a/a) λ = h/(mv)color(white)(a/a)|)))" "$ where $h = "Planck's constant" = 6.626 × 10^"-34"color(white)(l) "kg·m"^2"s"^"-1"$ $m =...

The idea here is that all matter can behave like waves, as given by the de Broglie hypothesis. The relationship between the wavelength of a massive particle, which is...

Louis de Broglie's key insight was that it is not just light that has a wave and a particle nature, but that other things such as electrons that we normally...

$sf(lambda=h/(mv))$ h is the Planck constant = $sf(6.63xx10^(-34)color(white)(x)Js)$ $sf(lambda)$ is the wavelength = 144 pm = $sf(1.44xx10^(-10)color(white)(x)m)$ $sf(m=1.67xx10^(-27)color(white)(x)kg)$ $:.$$sf(v=h/(mlambda)$ $sf(v=(6.63xx10^(-34))/(1.67xx10^(-27)xx1.44xx10^(-10))color(white)(x)"m/s")$ $sf(v=2.77xx10^(3)color(white)(x)"m/s")$

The formula for the de Broglie wavelength $λ$ is $color(blue)(bar(ul(|color(white)(a/a)λ = h/(mv)color(white)(a/a)|)))" "$ where $h =$ $m =$ the mass of the electron $v =$...

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

$lambda = 1.1 xx 10^(-38) "m"$ The de Broglie wavelength $lambda$ in $"m"$ is given by the de Broglie relation: $lambda = h/(mv)$ where: $h...

$lambda = "0.388 nm"$. Well, since an electron is a particle with mass, it can be described by the de Broglie relation: $lambda = h/(mv)$ where:...

You're actually looking for the de Broglie wavelength here, which, as you know, characterizes particles that have mass. The de Broglie wavelength depends on the momentum of the particle,...