In an electron microscope, electrons are accelerated to great velocities. What is the wavelength of an electron traveling with a velocity of #7.0 *10^3# #"km/s"# ? The mass of an electron is #9.1*10^(-28)# #"g"#.

1 Answer
Dec 6, 2017

Answer:

#1.0 * 10^(-10)# #"m"#

Explanation:

You're actually looking for the de Broglie wavelength here, which, as you know, characterizes particles that have mass.

The de Broglie wavelength depends on the momentum of the particle, i.e. the product between its mass and its velocity, as shown by the equation

#lamda_ "matter" = h/(m * v) #

Here

  • #lamda_ "matter"# is its de Broglie wavelength
  • #h# is Planck's constant, equal to #6.626 * 10^(-34)"J s"#
  • #m# is the mass of the particle
  • #v# is its velocity

In your case, you have

#{(m = 9.1 * 10^(-28)color(white)(.)color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 9.1 * 10^(-31)color(white)(.)"kg"), (v = 7.0 * 10^3 color(red)(cancel(color(black)("km")))"s"^(-1) * (10^3color(white)(.)"m")/(1color(red)(cancel(color(black)("km")))) = 7.0 * 10^6 color(white)(.)"m s"^(-1)) :}#

Now, you need to convert the mass of the electrons to kilograms and the velocity to meters per second because Planck's constant uses joules second as the unit.

Since

#"1 J" = "1 kg m"^2"s"^(-2)#

you can say that you have

#h = 6.626 * 10^(-34)color(white)(.)"kg m"^2"s"^(-1)#

So, plug in your values into the equation and solve for #lamda_("e"^(-))#, the de Broglie wavelength of the electron.

#lamda_ ("e"^(-)) = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg"))) "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(9.1 * 10^(-31)color(red)(cancel(color(black)("kg"))) * 7.0 * 10^(6) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))#

#lamda_ ("e"^(-)) = color(darkgreen)(ul(color(black)(1.0 * 10^(-10)color(white)(.)"m")))#

The answer is rounded to one significant figure, the number of sig figs you have for your values.