# Question 6ec4f

Nov 21, 2016

$9.90 \cdot {10}^{- 39} \text{m}$

#### Explanation:

In order to calculate the astronaut's matter wave, which is referred to as the de Broglie wavelength, you need to use

• her momentum, $p$
• Planck's constant, $h$, equal to $6.626 \cdot {10}^{- 34} {\text{kg m"^2"s}}^{- 1}$

The equation that gives you the de Broglie wavelength looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} = \frac{h}{p}}}} \to$ the de Broglie wavelength

Here

$p$ - the momentum of the astronaut
$l a m \mathrm{da}$ - her matter wavelength
$h$ - Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$

Now, the momentum of the astronaut is directly proportional to its velocity, $v$, which in your context can be taken to be its speed, and its mass, $m$

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{p = m \cdot v}}}$

Plug in your values to find

$p = {\text{201 kg" * "333 m s"^(-1) = "66,933 kg m s}}^{- 1}$

Now you're ready to calculate the de Broglie wavelength of the astronaut

lamda = (6.626 * 10^(-34)color(red)(cancel(color(black)("kg")))"m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^(-1)))))/("66,933" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("s"^(-1))))) = color(darkgreen)(ul(color(black)(9.90 * 10^(-39)"m")))#

The answer is rounded to three sig figs.