# Question 94aa0

Apr 5, 2017

$1.7 \cdot {10}^{- 10} \text{m}$

#### Explanation:

All you have to do here is to use the equation that establishes a relationship between the de Broglie wavelength of the electron$,$lamda$, \mathmr{and} i t s \ast m o m e n t u m \ast ,$p.

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

Here

• $p$ is the momentum of the electron
• $l a m \mathrm{da}$ is its de Broglie wavelength
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} {\text{kg m"^2"s}}^{- 1}$

The momentum of the electron depends on its mass and on its velocity as shown by the equation

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

Here

• $m$ is the mass of the electron
• $v$ is its velocity

before plugging in your values, make sure that the units match those used in the expression of Planck's constant.

Planck's constant uses the units ${\text{kg m"^2 "s}}^{- 1}$, which can be written as

$\text{kg m"^2 "s"^(-1) = color(blue)("kg m s"^(-1)) * "m}$

Notice that you have $\textcolor{b l u e}{\text{kg}}$ as the units for the mass of the electron and $\textcolor{b l u e}{{\text{m s}}^{- 1}}$ as the units for its velocity, so you can safely plug in the values as they were given and solve for $p$.

$p = 9.11 \cdot {10}^{- 31} \textcolor{w h i t e}{.} {\text{kg" * 4.3 * 10^6color(white)(.)"m s}}^{- 1}$

$p = 3.917 \cdot {10}^{- 24}$ ${\text{kg m s}}^{- 1}$

This means that the electron will have a de Broglie wavelength equal to

lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg m s"^(-1)))) * "m")/(3.917 * 10^(-24)color(red)(cancel(color(black)("kg m s"^(-1))))) = color(darkgreen)(ul(color(black)(1.7 * 10^(-10)color(white)(.)"m")))

The answer is rounded to two sig figs, the number of sig figs you have for the velocity of the electron.