# Video: Eg17S1-Physics-Q22

Eg17S1-Physics-Q22

04:09

### Video Transcript

The maximum intensity wavelength in the radiation from each of the Sun and a star π are 0.5 micrometers and 0.4 micrometers, respectively. Calculate the surface temperature of star π, given that the surface temperature of the Sun is 6000 kelvin.

Okay, so in this question, weβre receiving radiation from the Sun and a random star π. Weβve been told that, within the radiation spectra from each of these two stars, the maximum intensity wavelength occurs at 0.5 micrometers for the Sun and 0.4 micrometers for star π. In other words then, if we plotted a graph of radiation intensity on the vertical axis and the radiation wavelengths on the horizontal axis, then the radiation coming from the Sun would look something like this, where the wavelength where the intensity was maximum, thatβs this wavelength here, will be 0.5 micrometers. And if we plotted a similar curve for star π, then weβd see something like the pink curve, where the peak intensity wavelength was 0.4 micrometers.

Now, weβve been asked to calculate the surface temperature of star π, given that the surface temperature of the Sun is 6000 kelvin. Well, to answer this question, we can first of all assume that the Sun and star π are black bodies. Now, a black body is defined as any object that perfectly absorbs all the incident radiation without reflecting any. In other words, all of the radiation that falls on a black body gets absorbed by the black body. Now, the reason that we make this approximation is because first of all itβs a good one. The Sun and most stars can be approximated as black bodies. And secondly, this type of intensity wavelength distribution is a characteristic black body distribution. In other words, black bodies behave in this way, where they have a spectrum which shows a peak wavelength. And this peak wavelength depends on the temperature of the black body.

Specifically, we can say that the temperature of the black body, which weβll call π, is inversely proportional to the peak wavelength, which weβll call π. Or in other words, we can say that the temperature of the black body is equal to some constant of proportionality, which weβll call π, divided by the peak wavelength π. And it is important to remember that this π represents the peak wavelength, not just any wavelength.

So in the case of the Sun, letβs say that the temperature of the Sun or the temperature of the surface of the Sun more accurately, which weβll call π subscript π , is equal to that constant of proportionality divided by the peak wavelength of the Sun spectrum. Letβs call that π subscript s, s representing Sun of course. And similarly, for star π, we can say that its surface temperature, which weβll call π subscript π, is equal to the constant of proportionality once again, but this time divided by the peak wavelength in π spectrum, at which point we can take these two equations and divide one by the other.

So letβs say we divide this equation with this equation. On the left-hand side, weβd come out with π subscript π divided by π subscript π . And on the right-hand side, we have π over ππ, all divided by π over ππ , at which point weβve got a fraction divided by another fraction. And the way we can remember how to calculate that is to flip the dividing fraction and then multiply them together. So in other words, instead of dividing by π divided by ππ , we multiply by ππ  divided by π. And so we can see that the factors of π, the proportionality constant, cancel. And weβre just left with π π divided by π π  on the left-hand side and ππ  divided by ππ on the right, at which point we donβt even need to know anything about the constant of proportionality. All we need to know to find the surface temperature of the star π is the surface temperature of the Sun, which we already have, the peak wavelength of the Sun spectrum, which we also have, and the peak wavelength of the star π spectrum. Lucky us, we have that as well.

So letβs rearrange the equation to solve for π subscript π. We can do this by multiplying by π subscript π  on both sides of the equation and then seeing that π subscript π  cancels on the left-hand side. Then, what weβre left with is π subscript π is equal to ππ  over ππ multiplied by π subscript π , at which point we can plug in all the values weβve been given in the question. We then find that π subscript π is equal to 0.5 micrometers, the peak wavelength of the Sun spectrum, divided by 0.4 micrometers, the peak wavelength of the star π spectrum, multiplied by the surface temperature of the Sun, which is 6000 kelvin. And when we evaluate this whole fraction, we can see that the units of micrometers cancel in this fraction. And weβre gonna be left with the unit of kelvin, which is the unit of temperature. And this is brilliant cause thatβs what weβre finding.

And hence, we arrived at our final answer, which is that the surface temperature of star π is 7500 kelvin.