# What is the speed of a 50 g rock if its de Broglie wavelength is 3.32*10^-34 m?

## $h = 6.63 \cdot {10}^{-} 34 J \cdot s$

Aug 9, 2016

${\text{40 m s}}^{- 1}$

#### Explanation:

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 known as the de Broglie wavelength, and the momentum of the particle is described by the following equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} l a m \mathrm{da} \cdot p = h \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$l a m \mathrm{da}$ - the de Broglie wavelength
$p$ - the momentum of the particle
$h$ - Planck's constant, given to you as $6.63 \cdot {10}^{- 34} {\text{J s}}^{- 1}$

Your first goal here will be to use this equation to find the momentum of the particle. Rearrange to solve for $p$

$l a m \mathrm{da} \cdot p = h \implies p = \frac{h}{l a m \mathrm{da}}$

Before plugging in your values, use the fact that you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{1 J" = 1 "kg m"^2"s}}^{- 2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to write the Planck constant as

$h = 6.63 \cdot {10}^{- 34} \text{kg m"^2"s"^(-2) * "s}$

$h = 6.63 \cdot {10}^{- 34} {\text{kg m"^2"s}}^{- 1}$

Now plug in your values to find the momentum of the particle

p = (6.63 * 10^(-34)"kg m"^color(red)(cancel(color(black)(2))) "s"^(-1))/(3.32 * 10^(-34)color(red)(cancel(color(black)("m")))) = "1.997 kg m s"^(-1)

Now, the momentum of a particle is given by the product between its mass, $m$, and its velocity, $v$

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} p = m \cdot v \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Convert the mass of the particle from grams to kilograms

50 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.050 kg"

Plug in your values to find the velocity of the particle

$p = m \cdot v \implies v = \frac{p}{m}$

v = (1.997 color(red)(cancel(color(black)("kg")))"m s"^(-1))/(0.050color(red)(cancel(color(black)("kg")))) = "39.94 m s"^(-1)

Now, you don't have any information about the direction of motion for your particle, which means that you can say that the magnitude of the velocity, which is essentially the speed of the particle, is equal to

$v = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{40 m s}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer must be rounded to one sig fig, the number of sig figs you have for the mass of the particle.