### Video Transcript

A laser produces 4.00 times 10 to
the 27th photons, each with a frequency of 4.25 times 10 to the 14th hertz. What magnitude of momentum does
producing these photons impart on the laser? 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.

Okay, so in this example, we have a
laser. Let’s say that this is our
laser. And this Laser is producing
photons, which have a frequency — we’ll call it 𝑓 — of 4.25 times 10 to the 14th
hertz. Now, each one of our 4.00 times 10
to the 27th photons possesses some amount of momentum.

We can recall that that amount —
we’ll call it 𝑃 sub p, the momentum of a single photon — is equal to Planck’s
constant ℎ times the frequency of the photon divided by the speed of light in
vacuum. So here’s the idea. Each one of these photons as it
moves along has some momentum. And by the law of momentum
conservation, this same amount of momentum must be imparted in the opposite
direction on the laser. That is, as a photon is given this
amount of momentum to the right, it imparts this same amount of momentum in the
opposite direction on the laser.

If we add up the momentum of all
the photons, then we’ll get some much larger momentum amount. And once again, by momentum
conservation, this will be equal to the total amount of momentum imparted on the
laser.

Notice that our question asks us to
solve for the magnitude of momentum imparted. Which means that if we solve for
this amount, the total momentum possessed by all the photons, then we’ll have
answered our question. Because that’s equal to the
magnitude of momentum imparted on the laser. So how will we solve for this total
amount of momentum possessed by the photons emitted?

We can start by calculating the
momentum of each individual photon. We’ll call that 𝑃 sub p. And then realizing that all of the
photons that our laser emits have the same frequency. We can say that if capital 𝑁 is
the total number of photons emitted, then capital 𝑁 multiplied by 𝑃 sub p will be
equal to the total momentum of all the photons emitted. We’ll call that simply 𝑃.

This momentum, as we said, is equal
to the total momentum of the photons. But it’s also equal to the
magnitude of the momentum imparted to the laser. So in solving for 𝑃, we’ll find
the answer we want. We can start doing this by
replacing 𝑃 sub p, the momentum of an individual photon, with ℎ times 𝑓 divided by
𝑐, according to this equation. So the total momentum 𝑃 that we
want to calculate is equal to Planck’s constant. We’re told to treat this value as
equal to 6.63 times 10 to the negative 34th joule seconds. Multiplied by the frequency of each
photon 𝑓. This value is given as 4.25 times
10 to the 14th hertz. Divided by the speed of light in
vacuum 𝑐. We can approximate that value as
3.00 times 10 to the eighth meters per second. All multiplied by the total number
of photons emitted by our laser. That’s 4.00 times 10 to the
27th.

When we substitute all these values
in, before we go and calculate the total momentum 𝑃, let’s consider the units in
this expression. First, in our numerator, we see
that we have units of joules, which can be written equivalently as a newton times a
meter. And a newton, we can recall, can
also be written as a kilogram meter per second squared, which means that we can
replace a joule with a kilogram meter squared per second squared.

With that substitution made, let’s
now consider the units of hertz. These indicate number of cycles per
second and can be replaced with the units inverse seconds. With these replacements, we see
that some of our units can cancel out. In particular, in our units for the
Planck constant, we see that one factor of seconds can cancel out. And then along with this, one
factor of meters in our numerator and our denominator can cancel. And finally, one factor of inverse
seconds can cancel as well. This leaves us with the overall
units for this expression of kilograms meters per second. These are the units we expect for
momentum. So it looks like we’re on the right
track.

When we compute this fraction on
the left and keep three significant figures in our answer, we find a result of 3.76
kilograms meters per second. This is the total momentum of all
the photons emitted. And therefore, it’s the magnitude
of momentum the photons impart on the laser.