# Question #663b0

Dec 10, 2016

$4.8 \cdot {10}^{- 7} \text{m}$

#### Explanation:

In order to be able to calculate the de Broglie wavelength of the electron at that particular speed, you must know its mass

${m}_{\text{e" ~~ 9.1094 * 10^(-31)"kg}}$

Now, the de Broglie wavelength depends on the momentum of the electron, as given by the equation

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

Here

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

The momentum of the electron can be calculated by using 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 particle
• $v$ is its velocity

$p = 9.1094 \cdot {10}^{- 31} {\text{kg" * "1500 m s}}^{- 1}$

$p = 1.3664 \cdot {10}^{- 27} {\text{kg m s}}^{- 1}$

This means that the de Broglie wavelength associated with the electron at this particular speed is equal to

$l a m \mathrm{da} = \left(6.626 \cdot {10}^{- 34} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{kg"))) "m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^(-1)))))/(1.3664 * 10^(-27)color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s}}^{- 1}}}}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} = 4.8 \cdot {10}^{- 7} \text{m" = "480 nm}}}}$

The answer is rounded to two sig figs.