A popular kitchen appliance produces electromagnetic radiation with a frequency of 2450 MHz. With reference to Figure 6.4, answer the following: (c) If the radiation is not visible, do photons of this radiation have more or less energy than photons of visible light?

Video transcript

Hey everyone, we're told a certain medical procedure requires a patient to be temporarily exposed to radiation with a frequency of 27.12 MHz. Would the photons of this frequency have more or less energy than the photons of visible light. To answer this question, Let's go ahead and convert 27.12 MHz into Hertz. So we can easily compare it later on. Now, we know that per one MHz We have 10 to the six Hz. So this leaves us with 2.71, 2 times 10 to the seven Hz. And as we've learned, we know that our speed of light is going to be equal to our wavelength times our frequency. So let's go ahead and solve for our wavelength, which will be equal to our speed of light over our frequency. So, plugging in these values, We get 3.00 times 10 to the 8m/s, which is our speed of light. And we're going to divide that by our frequency of 2.712 times 10 to the seven hertz, which can also be written as second to the negative first. Now, when we calculate this out and cancel out our unit, we end up with 11.06 m which essentially tells us that we are in our radio waves. Now, why did we solve for our wavelength? Well, we solve for our wavelength because we know that our energy is inversely proportional to our wavelength. So since we have a larger wavelength than are visible light, this means that we have a smaller energy than our visible light. So to answer this question, Photons of 27.12 MHz have lesser energy than photons of visible light. And this is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.