Question #b44a1

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

`color(blue)(ul(color(black)(K_E = 1/2 * m* 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

`color(blue)(ul(color(black)(lamda = h/p))) ->` the de Broglie wavelength

Here

• `p` is the momentum of the electron
• `lamda` is its de Broglie wavelength
• `h` is Planck's constant, equal to `6.626 * 10^(-34)"kg m"^2"s"^(-1)`

The momentum of the electron can be expressed using the equation

`color(blue)(ul(color(black)(p = m * 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

`lamda = h/(m * v)`

Use the kinetic energy of the electron to

`K_E = 1/2 * m * v^2 implies v = sqrt( (2 * K_E)/m)`

Plug this into the above equation to find

`lamda = h/(m * sqrt( (2 * K_e)/m)) = h/sqrt( 2 * m * K_E)`

The mass of the electron is listed as

`m_"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)))`

`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)))))/(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)))))`

`lamda = color(darkgreen)(ul(color(black)(1.029 * 10^(-6)"m")))`

The answer is rounded to four sig figs.