Question

Question #459d4

Answer 1

`lamda = 2.4 * 10^(-11)"m"`

Explanation:

Start by looking up the mass of an electron

`m_"electron" ~~ 9.1094 * 10^(-31)` `"kg"`

Now, according to the de Broglie hypothesis, matter can actually behave like a wave.

In this case, your electron will exhibit wave-like behavior and have an associated matter wave. The wavelength of this electron is called the de Broglie wavelength and is 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)`

Notice that the equation uses the momentum of the electron, which depends on the velocity of the electron, `v`, and on its mass, `m`, as given by the equation

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

In your case, you have

`p = 9.1094 * 10^(-31)"kg" * 3 * 10^(7)"m s"^(-1)`

`p = 2.733 * 10^(-23)` `"kg m s"^(-1)`

Plug this into the equation that allows you to calculate the de Broglie wavelength to find

`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)))))/(2.733 * 10^(-23)color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))`

`color(darkgreen)(ul(color(black)(lamda = 2.4 * 10^(-11)color(white)(.)"m")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the velocity of the electron.