# Question #e8623

##### 1 Answer

#### Answer:

#### Explanation:

According to the **de Broglie hypothesis**, this electron can actually behave as a *wave*, in addition to its particle-like behavior.

This implies that your electron will have an associated **matter wave**. The *wavelength* of this matter wave, which is what we call the **de Broglie wavelength**, can be calculated using 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)#

As you know, the **momentum** of a particle depends on its **velocity**, **mass**,

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

In your case, you know that the electron is moving at

#v = 1/100 * 3 * 10^8# #"m s"^(-1)#

#v = 3 * 10^6# #"m s"^(-1)#

The mass of the electron is equal to

#m ~~ 9.1094 * 10^(-31)# #"kg"#

This means that the momentum of the electron is equal to

#p = 9.1094 * 10^(-31)color(white)(.)"kg" * 3 * 10^6color(white)(.)"m s"^(-1)#

#p = 2.733 * 10^(-24)# #"kg m s"^(-1)#

Plug this into the equation for 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^(-24)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^(-10)color(white)(.)"m")))#

I'll leave the answer rounded to two **sig figs**.