# Question Video: Comparing the Energy of Two Photons Physics • 9th Grade

In this question use a value of 6.63 × 10⁻³⁴ J⋅s for the Planck constant. A gamma ray typically has a frequency of around 3 × 10²⁰ Hz. Using this value, find the energy of a gamma-ray photon. Give your answer in scientific notation to two decimal places. A radio wave typically has a frequency of around 1.5 × 10⁶ Hz. Using this value, find the energy of a radio-wave photon. Give your answer in scientific notation to two decimal places. How many times more energetic is a gamma-ray photon of the given frequency than a radio-wave photon of the given frequency? Give your answer in scientific notation to two decimal places.

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### Video Transcript

In this question, use a value of 6.63 times 10 to the negative 34 joule-seconds for the Planck constant. A gamma ray typically has a frequency of around three times 10 to the 20 hertz. Using this value, find the energy of a gamma-ray photon. Give your answer in scientific notation to two decimal places.

To find the energy of a photon given its frequency, we recall a formula for quantized electromagnetic radiation that the energy of a photon is equal to the Planck constant times the frequency of the photon. We have a value for a Planck constant and also a value for the photon’s frequency. So all we need to do is multiply these values. We have 6.63 times 10 to the negative 34 joule-seconds times three times 10 to the 20 hertz. Now, we are looking for an energy but in addition to joules, which are units of energy on the right-hand side, we also have units of seconds and of hertz.

To remove seconds and hertz so that our overall units are joules, we can recall that one hertz is defined as one inverse second. This means that the units on the right-hand side of our expression for the energy are joulea-seconds per second. But seconds per second is just the dimensionless quantity one. So the overall units on the right-hand side are joules, which is exactly what we need for an energy. Multiplying 6.63 times 10 to the negative 34 by three times 10 to the 20, we get 1.989 times 10 to the negative 13 with our units of joules. Finally, rounded to two decimal places, 1.989 becomes 1.99. And the energy of our gamma-ray photon is 1.99 times 10 to the negative 13 joules.

It’s worth mentioning that we gave our answer to two decimal places. And this is consistent with the two decimal places that we are given for the Planck constant. However, this seems inconsistent with our value for the frequency, which is given with no decimal places. What’s going on is that we are told to use exactly this value in our calculations. In other words, this number is exact. So even though we haven’t written it with any decimal places, it doesn’t limit the precision of our final answer.

What makes the two numbers in this question different is that we are choosing to identify three times 10 to the 20 hertz as the typical frequency of a gamma-ray photon exactly. However, our value for the Planck constant is something that we determine from experiment, which means that it is inherently of limited precision.

Anyway, let’s move on to the second part of this question.

A radio wave typically has a frequency of around 1.5 times 10 to the sixth hertz. Using this value, find the energy of a radio-wave photon. Give your answer in scientific notation to two decimal places.

This question is essentially the same as the one we just answered. We’re looking for the energy of a photon given its frequency. This question is essentially identical to the one we just answered. The only difference is that we are using a different frequency, 1.5 times 10 to the sixth hertz, which is typical of radio waves. We also note again that this frequency is an exact value, so it won’t limit our final precision when we round to two decimal places. From our formula for the energy of a single photon, we know that all we need to do is multiply the given value for the Planck constant by the given frequency of the radio-wave photon.

As we know from before, we can make the units work out if we replace hertz with inverse seconds. So we have 6.63 times 10 to the negative 34 joule-seconds times 1.5 times 10 to the sixth per second. Seconds per second is just one. So we are left with joules, which is what we want because we’re looking for an energy. Performing the multiplication, we find that the energy of the photon is 9.945 times 10 to the negative 28 joules, which rounded to two decimal places is 9.95 times 10 to the negative 28 joules.