# Question b44a1

Feb 18, 2017

$1.029 \cdot {10}^{- 6} \text{m}$

#### Explanation:

The key here is the kinetic energy of the electron, which can be expressed as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{E} = \frac{1}{2} \cdot m \cdot {v}^{2}}}}$

Here

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

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 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 can be expressed 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 electron
• $v$ is its velocity

This means that the equation for the de Broglie wavelength can be rewritten as

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

Use the kinetic energy of the electron to

${K}_{E} = \frac{1}{2} \cdot m \cdot {v}^{2} \implies v = \sqrt{\frac{2 \cdot {K}_{E}}{m}}$

Plug this into the above equation to find

$l a m \mathrm{da} = \frac{h}{m \cdot \sqrt{\frac{2 \cdot {K}_{e}}{m}}} = \frac{h}{\sqrt{2 \cdot m \cdot {K}_{E}}}$

The mass of the electron is listed as

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

This means that the de Broglie wavelength of this electron is equal to

lamda = (6.626 * 10^(-34) "kg m"^2"s"^(-1))/(sqrt(2 * 9.1094 * 10^(-31)"kg" * 2.275 * 10^(-25)"kg m"^2"s"^(-2)))#

$l a m \mathrm{da} = \left(6.626 \cdot {10}^{- 34} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{kg"))) "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(sqrt(2 * 9.1094 * 10^(-31) * 2.275 * 10^(-25)) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s}}^{- 1}}}}\right)$

$l a m \mathrm{da} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.029 \cdot {10}^{- 6} \text{m}}}}$

The answer is rounded to four sig figs.