Question #459d4

1 Answer
Apr 16, 2017

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.