Video: Calculating the Magnitude of Laser Photons’ Momentum

A laser produces 4.00 × 10²⁷ photons, each with a frequency of 4.25 × 10¹⁴ Hz. What magnitude of momentum does producing these photons impart on the laser? Use a value of 6.63 × 10⁻³⁴ J⋅s for the Planck constant. Give your answer to 3 significant figures.

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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.

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