"The diameter of a rubidium atom is 495 pm We will consider
two d...

Question

"The diameter of a rubidium atom is 495 pm We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms:

(c) If extended to three dimensions, which arrangement would lead to a greater density for Rb metal?"

Accepted Answer

Welcome back everyone in this example we're told the diameter of potassium atom is 235 kilometers. And shown below are two different ways to pack the atoms. So we have a one dimensional arrangement and we have a two dimensional arrangement. And according to the promise, we need to identify which arrangement would lead to a greater density for potassium if the packing is extended to three D. Dimension. So We know according to the prompt, the diameter of each of our potassium atoms is sorry 235 m. So we want to now that we know that that's our diameter. We want to assume a side length and we'll say not of but we'll just say equal to one m. So we'll draw that out And this is going to represent one m of distance. So we want to figure out how many of our potassium atoms would be equal or would fit in a row Of a side length of one m. So we would say number of Kay Adams we'll say aligned in the side of one or side equal to one m, we can say so we need to figure out the amount of atoms that can fit In a side length of one m in a row. So what we're going to do Is now that we know our diameter and we assume that we have a side length of one m. We're going to do a dimensional analysis step. So this is all for part one Or for arrangement one and we can say that because we assume we have a side length of one m. And we know according to the prompt that we have a diameter of m per potassium adam. We want to get rid of these units. So we're going to multiply by a conversion factor. We're in our numerator. We want our PICO meter units. So we call that PICO meter, tells us that we have 10 to the 12th power PICO meters for our base unit one m. And so this allows us to cancel out meters as well as PICO meters, leaving us with just our assumed unit being atoms here. And this is going to give us a value equal to 4.256 times 10 to the ninth power atoms. Now this is specifically the number of atoms that can fit in a row Aligned to a one m distance. But as we see an arrangement one, we have a square arrangement here. So I'll just outline our square here. We have a square arrangement of our atoms. So now we want to go ahead and continue this calculation to calculate the number of potassium atoms for the square that can be aligned to a side length of one m. So we want to recall actually our area of a square. So recall that that would be area equal to each side squared. And because above we calculated the amount of atoms that can fit in a side length of one m, we would say that our area is equal to. And we're going to plug in for side our number of atoms. So above we stated that that is 4.256 times 10 to the ninth power. And this is going to be squared. And this is going to give us a value equal to 1.812 times 10 to the 19th power atoms. So gives this gives us an idea of the density of our atoms that are packed in arrangement one. Now that we know the area of the square, an arrangement one. So we want to follow the same process and compare this to arrangement too. So we'll do the work down there for blow. And what we should recognize is that an arrangement to? We have a series of triangular arrangements of our potassium atoms because we now have an offset square. And as you can see the gaps from arrangement 12 now arrangement to between our atoms is now much smaller in arrangement too. And we should recognize that the triangle or the triangular arrangement that we can identify an arrangement to leads us to a right triangle where in the middle we have our angle by sector here where we have our right triangle, 90° angle. And we want to recognize or rather recall that to calculate the different sides for a right triangle. We have our diameter below which represents the entire side here, we have our height of our right triangle represented in the middle by our angle by sector being two times height. And then we have this third side here which is going to be two times the diameter. However, the most important side is going to be again two H, which represents our angle by sector. And we want to pay attention to this side because this is going to be two times our vertical distance that is occupied by a row of atoms. So just to write that below two H is going to be the vertical distance occupied by the row of atoms. And so this means that therefore we want to solve for height. And so because we have a right triangle, we want to go ahead and recall the pythagorean theory. Um So I'll write that below. So we would recall that our pythagorean theorem tells us for a right triangle. We have the relationship where we have two times height squared plus the diameter squared set equal to two times the diameter squared. And so to isolate for height, we would have height squared here and we would need to get rid of that square By using a fraction to the 1/ power. And because we have two times two, that gives us a denominator of four, so so far we can say height is equal to we would have our like terms here. So D squared added to two D squared, would give us three D squared. So that would be our numerator, three D squared. And then as we stated, to get to square we have four. So we would divide by four and as we stated because we have this square root attached to our height. We would cancel it out by using the fraction one half. So we would take this quotient and raise it to the one half power. And so this is how we would isolate for height. And so now we want to solve for height plugging in the info that we know. So we would get three times our value for D which is going to be our diameter according to the prompt. Given us 235 centimeters. This is squared and then divided by r value in the dominator four and raised to the one half power. And so what we're going to get is a value for height equal to 203. m. And so now that we have this value for height as we followed in part one of our solution for depiction one, we used the calculation for side but we're going to use height for part two to calculate for in place of side. So we want to find the number of rows or the number of atoms in a row For a side equal to one m. But in this case we know that our height is now going to be the value that we calculated above. So what we would have and I'll make more room here is we're assuming we have one m of length to fill with our potassium atoms and we would divide by our calculation above, which is our height being 203.516 PK meters. And we're going to multiply by the relationship to cancel out our units so that we end up with atoms as our final unit. By recalling that are prefixed, PICO tells us that we have 10 to the 12 power meters equivalent to one m of our base unit meter. And so we can cancel out our units here. We can get rid of meters. We can get rid of parameters. And what we're going to get here is a value equal to 4.914 times 10 to the ninth power. And just to make a correction here, this should be for our number of rows, actually for a side. So a number of rows of potassium atoms for a side equal to one m. And that is due to the fact that we solved for the height as this value here, which as we stated, is our vertical distance occupied by a row of atoms. And so that is why instead our final value here represents our rose that we should have of our potassium atom in one m of a side length. And so now that we have this value, we're going to scroll down for more room. And as we did above, we want to calculate for the square arrangement. And so we're going to take our value which we found above which is our number of atoms in the square which in part one we saw for. And that was that was actually not this value for area but this value here Which is our atoms that can fit in a row aligned to a one m side length. So that was 4.25, 6 times 10 to the 9th power atoms. And we want to go ahead and multiply this buyer value for the number of rows that would fit for a side of one m and above. We calculated that as 4.914 times 10 to the 9th power rose. And so what we're gonna get here is a value that is equal two And we'll use a different color here. We'll just use black 2.091 times 10 to the 19th power atoms. And so now that we have this value, we want to get the ratio of atoms between our two arrangements. So above we found that ratio by our area value here, which was this value that we found. And now we have this value here for the square in part two or an arrangement to And so for our ratio of atoms we would see that we have the higher value in the numerator. So we'll place 2.91 times 10 to the 19th power atoms divided by our smaller value in the denominator 1.812 times 10 to the 19th power atoms. And this is going to give us a ratio that is equal to about we'll say about 1.15 times. And so that means that we can say that therefore not only does arrangement too have more atoms per square unit, but arrangement three, We can say but arrangement three Or we can say arrangement in three D. Because we are not shown arrangement three would lead to more atoms per cubic unit than arrangement to. And so we would say that for our final answer. We can say thus and I'll make more room arrangement too leads two will say a greater density of potassium atoms. And so our final answer is simply just going to be arrangement to. And as we stated, that is because we calculated that we have a ratio where arrangement two is 1.15 times that of arrangement one. And that would make sense because as we stated, the gaps between each of our atoms and arrangement to is a much smaller space compared to the gaps between our atoms and arrangement one. And so we have a lot more density in arrangement too. So hopefully what I've explained helps what's highlighted in yellow here is our final answer. And if you have any questions just leave them down below. So I will see everyone in the next practice video.

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