# Question #459d4

##### 1 Answer

#### Answer:

#### 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))) -># thede 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,

#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.