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Question Video: Finding the Frequency of a Photon Given Its Momentum Physics • Third Year of Secondary School

A photon has a momentum of 1.11 × 10⁻³³ kg⋅m/s. What is the frequency of the photon? Use a value of 6.63 × 10⁻³⁴ J⋅s for the Planck constant. Give your answer to the nearest megahertz.

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Video Transcript

A photon has a momentum of 1.11 times 10 to the negative 33 kilogram meters per second. What is the frequency of the photon? Use a value of 6.63 times 10 to the negative 34 joule-seconds for the Planck constant. Give your answer to the nearest megahertz.

To answer this question, we will need to calculate the frequency of a photon knowing its momentum and then report this frequency in units of megahertz.

The momentum of a photon can be expressed as 𝑃, the photon’s momentum, is equal to ℎ, the Planck constant, times 𝑓, the photon’s frequency, divided by 𝑐, the speed of light in vacuum. Since we know the momentum and are trying to find the frequency, let’s rearrange this formula by multiplying both sides by 𝑐 divided by ℎ. On the left-hand side, we just have momentum times speed of light divided by the Planck constant. On the right-hand side, 𝑐 divided by 𝑐 is one and ℎ divided by ℎ is also one, so we are just left with frequency.

We are given a value for the photon’s momentum and also a value for the Planck constant. So the last thing that we need is a value for the speed of light in vacuum. For the purposes of this question, it is sufficient to use a value for the speed of light of 3.00 times 10 to the eighth meters per second. This value is always a good starting point for the speed of light. And if we go through the calculation and discover that our result is not sufficiently accurate, all we need to do is perform the same calculation again but with a more accurate value for the speed of light.

Nevertheless, for this particular question, we already know that two decimal places is sufficient because both of the values we’re given in the question are only accurate to two decimal places, which means two decimal places of accuracy is all we need.

Substituting values, we have 1.11 times 10 to the negative 33 kilogram meters per second times 3.00 times 10 to the eighth meters per second divided by 6.63 times 10 to the negative 34 joule-seconds. Kilogram meters per second times meters per second is kilograms meter squared per second squared, which is exactly one joule. So the units in the numerator are joules, and the units in the denominator are joules times seconds. Joules divided by joule-seconds is just inverse seconds, which are hertz. So the overall units of this quantity is hertz, which tells us we are on the right track because hertz is a unit for frequency.

Evaluating the numerical portion of this calculation gives us 502.262 and several more decimal places times 10 to the sixth with units of hertz. Now remember we want our answer in units of megahertz. So we recall that one megahertz is equal to 10 to the sixth hertz. Looking back at our answer, we see that it is already expressed as a mantissa times 10 to the sixth hertz. So we can just replace 10 to the sixth hertz with megahertz to get our answer in appropriate units.

Finally, we want our answer to the nearest megahertz, which means rounding 502.262 et cetera to the nearest integer. Well, the nearest integer to 502.262 et cetera is just 502. So the frequency of our photon is 502 megahertz. This frequency happens to be part of the ultra high frequency radio wave portion of the electromagnetic spectrum.

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