### Video Transcript

What is the de Broglie wavelength
of an electron that has a momentum of 4.56 times 10 to the negative 27th kilograms
meters per second? Use a value of 6.63 times 10 to the
negative 34th joule-seconds for the Planck constant. Give your answer to three
significant figures.

All right, so in this example, we
have an electron. And an electron is moving along
with some speed. Since the electron has mass and is
in motion, that means it has some amount of momentum. And we’re told just how much that
amount is. So knowing the mass of an electron,
we could work back, if we wanted to, to solve for its velocity. But rather, what we want to do is
calculate the de Broglie wavelength of this electron. To do that, we can recall that this
wavelength, we’ll call it 𝜆, is equal to Planck’s constant ℎ divided by 𝑝, the
momentum of the object. Using a value of 6.63 times 10 to
the negative 34th joule-seconds for ℎ, we’re able to use that value along with the
momentum 𝑝 of our electron to solve for its de Broglie wavelength. So then, here we have our equation
for the de Broglie wavelength of the electron. But before we calculate this
fraction, let’s consider the units in this expression.

We see that in the numerator, we
have units of joule-seconds. A joule is equal to a newton times
a meter. And a newton is equal to a kilogram
times a meter divided by a second squared, which means we can replace the joule in
our units with kilograms meter squared per second squared. When we do this, we now see that we
can cancel out some of these units. In the numerator, one factor of
seconds will cancel out. And then we can see that both
numerator and denominator have a factor of one over seconds. So that cancels as well. And along with this, our units of
kilograms cancel one another out, as does one factor of meters.

In the end, the only unit left is a
single factor of meters. That’s a good sign because we’re
calculating a distance, a wavelength. With each one of the values in our
fraction given to three significant figures, we also want to give our answer to that
precision. That works out to be 1.45 times 10
to the negative seventh meters. Expressed in units that may be more
familiar, this is equal to 145 nanometers. That’s the de Broglie wavelength of
this electron.