### Video Transcript

The following graph shows the
intensity of light at different wavelengths for a radiating blackbody at two
different temperatures. Within which of the regions shown
is there a greater difference in the number of photons of a given frequency emitted
at the greater temperature compared to that emitted at the lesser temperature? (A) Region 1, (B) region 2, (C) the
difference in the number of photons emitted at a given frequency is the same in both
regions.

In this question, we are asked to
determine the region of the graph in which the difference in the number of photons
emitted for a given frequency is greater for a blackbody at two different
temperatures. Letโs look at the graph.

We can see that as the wavelength
increases along the axis, the intensity of light initially increases for each
temperature, until a peak intensity is reached. The wavelength at these particular
intensities weโve labeled here as ๐ max. The intensity then decreases at a
lower rate, creating a skewed tail to the right of the peak. Their slopes are lower. After the peak intensities, the
lines then get closer together.

We can also observe that if we
choose a wavelength within region 1 and another wavelength within region 2, then the
difference in intensity of light at each temperature, and therefore in energy
emitted as well, is greater in region 1 than in region 2. In order to determine which region,
1 or 2, has a greater difference in the number of emitted photons of a given
frequency between the two temperatures, we first have to determine how exactly we
can compare the number of photons.

To do this, letโs recall Planckโs
quantization equation. The total energy of all of the
photons, ๐ธ ๐, is equal to ๐โ๐, where โ is Planckโs constant, ๐ is the frequency
of the photons, and ๐ is the total number of photons. Note that because you canโt have
half of a photon, ๐ must always be a whole integer.

Let us also remember that the speed
of light, ๐, is equal to the frequency of light times its wavelength. Or in terms of frequency, ๐ is
equal to ๐ divided by ๐. With this relation, we can rewrite
the energy equation like so. ๐ธ ๐ is equal to ๐โ๐ divided by
๐.

Now then, looking back at the
graph, we can choose the wavelength ๐ at the right end of each region to look at,
which weโll call ๐ one and ๐ two. We then determine the difference in
energies in each region and, from observation, verify that the intensity of light
for the red curve, that is, the one with the highest temperature, is greater and its
energy is greater than for the blue curve for the lower temperature. This is true in both regions 1 and
2. But in region 1, these energy
differences are greater than in region 2.

Now, letโs remember that both โ and
๐ are constant and that we are comparing energy differences with the same
wavelength ๐ within each specific part of the region, the right side. This means, looking back at the
Planck equation, all of our variables are the same when we look at the right side of
the region for both regions 1 and 2, except for the number of photons ๐. Thus, we must conclude that the
difference between the number of quanta of light for these frequencies is what is
responsible for the difference in energy in these regions.

Knowing that in region 1 the
difference between the curves is greater than in region 2, where they are closer,
this must mean the difference of the integers ๐ in region 1 must be greater than in
region 2. So our correct answer must be
option one, in region 1.