### Video Transcript

What is the difference in the energy of a blue photon with a wavelength of 400 nanometers and a red photon with a wavelength of 700 nanometers? Use a value of 6.63 times 10 to the negative 34 joule-seconds for the value of the Planck constant and 3.00 times 10 to the eighth meters per second for the value of the speed of light in free space. Give your answer in joules in scientific notation to two decimal places.

To answer this question, we will need to know how to relate the energy of a photon to its wavelength. It’s worth mentioning that when we speak about photon color — in this question, we have a blue photon and a red photon — we mean something different than when we talk about the color of a physical object. The color of a photon is another way to characterize its wavelength. 400 nanometers is in the blue portion of the visible light spectrum, so we call a photon with this wavelength a blue photon. Similarly, 700 nanometers is in the red portion of the visible light spectrum, so we call a photon with this wavelength a red photon. On the other hand, the color of an object is another way to describe which wavelengths of photons are most strongly reflected or transmitted by that object.

Getting back to our question, the energy of a photon can be expressed in terms of the Planck constant, the speed of light in free space, and the photon’s wavelength as 𝐸, the energy of the photon, is equal to ℎ, the Planck constant, times 𝑐, the speed of light in free space, divided by 𝜆, the wavelength of the photon. Now, we are looking for the energy difference between two photons. To do this, we simply subtract the energy of one photon from the energy of the other. If we use the subscripts 𝑏 and 𝑟 to represent the blue and red photons, respectively, we can write 𝐸 sub 𝑏 minus 𝐸 sub 𝑟 as the energy difference we are looking for.

Now, we substitute our previous expression. We have ℎ𝑐 divided by 𝜆 𝑏 minus ℎ𝑐 divided by 𝜆 𝑟. Note that in both of these terms the value of ℎ𝑐 is the same because the Planck constant and the speed of light in free space are both constants. But 𝜆 𝑏 and 𝜆 𝑟 are different because we have to use the specific wavelength for each photon. At this point, we could just substitute values and calculate. But before that, let’s factor the constant ℎ𝑐 out of both terms. We’ll do this because factoring out constants can often help us gain physical insight into an expression. This gives us ℎ𝑐 times the reciprocal of 𝜆 𝑏 minus the reciprocal of 𝜆 𝑟.

In this form, we can make the physical observations that the energy difference between two photons depends on the reciprocals of their wavelengths and also that the scale of this difference is set by the constant value of the Planck constant times the speed of light in free space. Now, all that’s left is to substitute values. Note that our value for the Planck constant is given in scientific notation and SI base units. Likewise, the speed of light in free space is also given in scientific notation with SI base units. So, to make our calculations easier, let’s express the wavelength of the photons, 400 nanometers and 700 nanometers, in scientific notation with SI base units.

Recall that one nanometer is 10 to the negative nine meters. So 𝜆 𝑏 is 400 times 10 to the negative nine meters, and 𝜆 𝑟 is 700 times 10 to the negative nine meters. Substituting these wavelengths and our value for the Planck constant and the speed of light in free space into our expression, we have 6.63 times 10 to the negative 34 joule-seconds times 3.00 times 10 to the eighth meters per second times the quantity one divided by 400 times 10 to the negative nine meters minus one divided by 700 times 10 to the negative nine meters. If we evaluate the term in parentheses, the difference between the reciprocals of the wavelengths, we get 1.07 times 10 to the sixth inverse meters. Note that inverse meters and one over meters are the same unit. Multiplying these three numbers together gives us 2.12823 times 10 to the negative 19 joule-seconds meters per second inverse meters.

Let’s simplify these units. Seconds per second is just one, so we can get rid of those. And meters times inverse meters is also one, so we can get rid of those as well, and the overall units are joules. This is good because the question asked us to give our answer in joules. The question also asked us to give our answer in scientific notation, and our current answer is indeed in scientific notation, and also to two decimal places. So we need to round 2.12823 to two decimal places. The rounded number is just 2.13, and our full final answer with units of joules expressed in scientific notation to two decimal places is 2.13 times 10 to the negative 19 joules.

As a side note, blue and red are the two ends of the visible light spectrum. So what the answer to this question tells us is that the visible light spectrum has a width of approximately 2.13 times 10 to the negative 19 joules or about 1.3 electron volts.