# 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

#### Answer:

#### 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# isPlanck'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"^(-)) = (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.