# data[].attributes.attachment_link.title
Stefan Boltzmann law and Emissivity | video in HINDI

# data[].attributes.attachment_link.description
In this Physics video lecture in Hindi on Radiation for class 11 we explained Stefan Boltzmann law of blackbody radiation. During the derivation of the equation for the rate of radiation for any surface we found that it is equal to the product of emissivity of the surface, the Stefan's constant, the surface area and the fourth power of the absolute temperature of the object.

# data[].attributes.attachment_link.readable_text
- [Instructor] In a previous video, we introduced ourselves to the idea of average atomic mass, which we began to realize could be a very useful way of thinking about a mass at an atomic level, or at a molecular level. But, what we're gonna do in this video is connect it to the masses that we might actually see in a chemistry lab. You're very unlikely to just be dealing with one atom, or just a few atoms, or just a few molecules. You're more likely to deal with several grams of an actual substance. So, how do we go from the masses at an atomic scale to the masses, masses of samples that you see in an actual chemistry lab, or in, I guess you could say, r-scale of the world. Well, the chemistry community has come up with a useful tool. They said, all right, let's think about a given element. So, say, lithium. We know its average atomic mass is 6.94, 6.94 unified atomic mass units per atom, atom of lithium. What if there were a certain number of atoms of lithium such that if I have that number, so times certain, certain number of atoms, then I will actually end up with 6.94 grams, grams of lithium. And, this number of atoms is 6.02214076 times 10 to the 23rd power. So, if you have a sample with this number of lithium atoms, that sample is going to have a mass of 6.94 grams. Whatever its average atomic mass is in terms of unified atomic mass units, if you have that number of the atom, you will have a mass of that same number in terms of grams. Now, you might be saying, is there a name for this number, and there is indeed a name, and it is called Avogadro's number, named in honor of the early 19th century Italian chemist, Amedeo Avogadro. And, in most contexts, because you're not normally dealing with data with this many significant digits, we will usually approximate it as 6.022 times 10 to the 23rd power. Now, there's another word that it's very useful to familiarize yourself with in chemistry, and that's the idea of a mole. Now, what is a mole? It is not a little mark on your cheek. It is not a burrowing animal. Actually, it is both of those things, but, in a chemistry context, a mole is just saying you have this much of something. The word mole was first used by the German chemist Wilhelm Ostwald at the end of the 19th century, and he came up with the word because of its relation to molecule. Now, what does that mean? Well, think about the word dozen. If I say I've got a dozen of eggs, how many eggs do I have? Well, if I have a dozen of eggs, that means I have 12 eggs. So, if I say I have a mole of lithium atoms, how many lithium atoms do I have? That means that I have 6.02214076 times 10 the to 23rd lithium atoms. Exact same idea, it's just that Avogadro's number is much hairier of a number than a dozen. So, let's use our new found powers of the mole and Avogadro's number to start doing some useful things. Let's say that someone were to walk up to you and say, hey, you, I have a 15.4 milligram sample of germanium. How many atoms of germanium am I dealing with? Pause this video and try to think about that. So, let me clear out some space the periodic table of elements was taking up. All right, so we started off with 15.4 milligrams of germanium. The first step might be hey, let's convert this to grams of germanium. And so, we can do a little bit of dimensional analysis. We can just multiply this, for every one gram of germanium that is equivalent to 1,000 milligrams, milligrams of germanium. And so, if you essentially multiply by one thousandth or divide by 1,000, we're gonna get the grams of germanium. And, you can see that in the dimensional analysis by seeing that that is going to cancel out with that leaving us with just the grams of germanium. And, now that we have an expression for grams of germanium, we can think about moles of germanium. So, how do we do that? Well, we're going to multiply by some quantity, and in the denominator we're going to want grams of germanium for the dimensional analysis to work out, grams of germanium, and in the numerator we want the new expression to be in terms of moles of germanium. So, one mole of germanium is equal to how many grams of germanium? Well, we see it right over here. Germanium's molar mass is 72.63 grams per mole. So, for every mole, we have 72.63 grams of germanium. And, you can see that the units work out. These grams of germanium are going to cancel with the grams of germanium just leaving us with moles of germanium. In an actual chemistry practice, finding out the moles of a substance might actually be the most useful thing, but if you wanted to find out the actual atoms of germanium that we're dealing with, we will just multiply by the number of atoms you have per mole. And, this is going to be true for any element. For every mole, you have Avogadro's number of atoms. And, we're going to approximate that as 6.022 times 10 to the 23rd atoms, atoms of germanium, for every one mole, mole of germanium. And so, just to review what we just did, we had milligrams of germanium. You multiply these two together, you'll have grams of germanium, which makes sense, you're essentially just dividing by 1,000. If you were to multiply your grams of germanium times the moles per gram, which is really just the reciprocal of this molar mass we got here, and just to make sure where it makes sense, the units work out nice with the dimensional analysis, this right over here tells you your moles, moles of germanium. And then, if you take your moles and then you multiply it by Avogadro's number, it tells you how many atoms of germanium we have, and that makes sense. If I told you I had a certain number of dozen of eggs, if I wanted to know how many eggs that is I would multiply by 12. So, this whole expression is the number of atoms, atoms of germanium. So, we have 15.4 milligrams. If we wanna figure out how many grams we have, we then divide by 1,000, that's what our dimensional analysis tells us, and it also makes logical sense, divided by 1,000. So, this is how many grams we have. And then, if we wanna figure out how many moles, and it's going to be a small fraction of a mole because a mole is 72.63 grams per mole, we have a small fraction of a gram, much less 72.63 grams. And so, we saw from our analysis to figure out the number of moles, we're now going to essentially divide by 72.63, so divided by 72.63 is equal to, this is the number of moles of germanium we have. And, if we wanna figure out the number of atoms of germanium, we'll then multiply that times Avogadro's number. So, times 6.022 times 10 to the 23rd, and this EE button means times-10-to-the, EE 23rd power, so that's how you do it on a calculator. And then, that gives us this many atoms. And, let's see, just to get our significant digits here, our significant figures, out of all of the things we multiplied, see we had four significant digits here, four significant digits here, but we only had three over here, so I'm going to round to three significant digits. So, I'll go to 1.28 times 10 to the 20th atoms. So, we have approximately 1.28 times 10 to the 20th atoms of germanium, which is a lot.

# data[].attributes.attachment_link.title
The mole and Avogadro's number (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Boyle's law (video) | Ideal gas equation | Khan Academy

# data[].attributes.attachment_link.description
robert boyle was an Irish scientist in the 1600s in it and it's actually from his experiments that we get Boyle's law which actually preceded the ideal gas equation and we've already shown that but we're going to work backwards and we'll use Boyle's law to prove part of the

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.description
In this video, David explains what a photon is and how to determine the energy of a photon.

# data[].attributes.headline
Thermal conduction, convection, and radiation

# data[].attributes.subject
Thermal conduction, convection, and radiation

# data[].attributes.attachment_link.title
Thermal conduction, convection, and radiation (video) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.attachment_link.title
Properties of Light: Blackbody Radiation

# data[].attributes.attachment_link.description
A look at the property where, at high temperatures, objects will begin to glow and the spectrum of the light they emit will depend on temperature. This can be used to measure the temperature of distant objects. Part of an introductory series on astronomy Let us know what you think of these videos by filling out our short survey at http://tinyurl.com/astronomy-pulsar.