Question #94aa0

1 Answer
Apr 5, 2017

Answer:

#1.7 * 10^(-10)"m"#

Explanation:

All you have to do here is to use the equation that establishes a relationship between the de Broglie wavelength of the electron#, #lamda#, and its **momentum**, #p#.

#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)#

The momentum of the electron depends on its mass and on its velocity as shown by the equation

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

Here

  • #m# is the mass of the electron
  • #v# is its velocity

before plugging in your values, make sure that the units match those used in the expression of Planck's constant.

Planck's constant uses the units #"kg m"^2 "s"^(-1)#, which can be written as

#"kg m"^2 "s"^(-1) = color(blue)("kg m s"^(-1)) * "m"#

Notice that you have #color(blue)("kg")# as the units for the mass of the electron and #color(blue)("m s"^(-1))# as the units for its velocity, so you can safely plug in the values as they were given and solve for #p#.

#p = 9.11 * 10^(-31)color(white)(.)"kg" * 4.3 * 10^6color(white)(.)"m s"^(-1)#

#p = 3.917 * 10^(-24)# #"kg m s"^(-1)#

This means that the electron will have a de Broglie wavelength equal to

#lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg m s"^(-1)))) * "m")/(3.917 * 10^(-24)color(red)(cancel(color(black)("kg m s"^(-1))))) = color(darkgreen)(ul(color(black)(1.7 * 10^(-10)color(white)(.)"m")))#

The answer is rounded to two sig figs, the number of sig figs you have for the velocity of the electron.