# Question #663b0

##### 1 Answer

#### Explanation:

In order to be able to calculate the **de Broglie wavelength** of the electron at that particular speed, you must know its mass

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

Now, the de Broglie wavelength depends on the **momentum** of the electron, as 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 wavelength#h# - Planck's constant, equal to#6.626 * 10^(-34)"kg m"^2"s"^(-1)#

The momentum of the electron can be calculated by using the equation

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

Here

#m# is the mass of the particle#v# is its velocity

In your case, you have

#p = 9.1094 * 10^(-31)"kg" * "1500 m s"^(-1) #

#p = 1.3664 * 10^(-27)"kg m s"^(-1)#

This means that the de Broglie wavelength associated with the electron at this particular speed is equal to

#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)))))/(1.3664 * 10^(-27)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 = 4.8 * 10^(-7)"m" = "480 nm")))#

The answer is rounded to two **sig figs**.