"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) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A?

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) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A?

Comparison of two atomic arrangements: (i) square grid and (ii) close-packed arrangement.

(c) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A?

Accepted Answer

Welcome back everyone. We're told that the diameter of a atom of potassium is 235 km. And shown below are two ways to pack the atoms. We need to determine how has the number of atoms on the surface increased. Going from arrangement one to arrangement too. So according to the prompt were given our diameter for each atom of potassium as 235 m. And what we wanna do is make an assumption here. So we want to assume a length of one m And we want to see how many of our potassium atoms can line up in a row to the length of one m. So we want to measure the side by side distance or rather the side by side amount of potassium atoms that can be aligned in a row for a distance of one m. So beginning with part or arrangement one we can say we are told from the prompt that we have a diameter of 235 PK meters. And we're going to relate this to our distance that we're measuring being one m. So we would say that we have one m divided by the diameter given the prompt as 235 m. And we want to go ahead and multiply in a dimensional analysis step by a conversion factor where to cancel our units of PICO meters. We want that in the numerator and recall that our prefix PICO tells us that we have 10 to the 12th power of our prefix unit here, which corresponds to one of our base unit meter. And so this allows us to cancel meters at the same time as PK meters. And what we're going to be left with is our implied unit, which just leaves us with our atoms of potassium. And this is going to equal a value of 4.256 times 10 to the ninth power atoms. And these are going to be the atoms that fit in a row Measured to a length of one m. However, we should recognize that our arrangement in part one is a square arrangement here. And so we want to recall the area of a square. And so to continue our calculation, we would recall that our area of a square is equal to the side squared. So because we know from our calculation above that this amount of atoms would make up one m of side length for each side of our square. We can incorporate that into our calculation for area. And so we can say that our area is equal to in brackets. We would plug in 4.756 times 10 to the ninth power atoms. We have that squared and this is going to give us an area value that is equal. And just to make a correction here, this should be 256 in both of our steps. So when we square this value, we're going to get a value that is equal to 1.811 times 10 to the 19th power atoms. And this would be our number of atoms packed into the density of our square or the area of our square, which can also give us an idea of its density. So now looking at our arrangement in arrangement to we can see that based on our atoms, we have a triangular arrangement that make up the offset square in this molecule. And because I'm sorry, let's draw better triangle here. So because we have these right triangles that we can identify here, we want to recall that in a right triangle. We have our angle by sector here, which divides not only at the bottom side of our triangle, which is our diameter where we have two new sides where the side for our or the side in the middle of our triangle is going to be the side characterized as two times the height where the third side of our triangle. This side here in purple can be characterized by two times our diameter. And so the most important side here is going to be two times the height, which is going to be something we recall as our angle by sector. And we should recall that the angle by sector is important because it's two times the vertical distance that is occupied by a row of atoms. So moving on in our solution below, we're going to compare the area of value that we calculated above with what we can get for our arrangement in part or an arrangement to here. So because we're dealing with a right triangle, we want to recall our pythagorean theorem and we would recall that that is telling us that we have two times height squared added to the diameter squared, which should equal two times the diameter, which is then squared again. And so we want to solve for our vertical distance and that would be for height. So to solve for height, we would isolate for H. And in doing so, we would say that height is equal to we would have three times the distance squared because we would combine our like terms where we have one D squared plus two D squared here. That gives us three D squared. And then we would be dividing by four. And that comes from the two which is squared here. And then because we need to isolate for height. We would raise this entire quotient to a power of one half to cancel out our square there. So raising this to a power of one half and that would be our complete steps for isolating height. So now we can just plug in our value for distance. And so we would use the or sorry, not for distance, but for D as in diameter. And so we would use our diameter given in the prompt. So we would have three times our diameter in the prompt of 325 PK meters. This is squared divided by four and raised to the one half power and in our calculators, we should get a value equal to 203.516 PK meters here. And so this value represents the distance occupied by a row of atoms. And so now we want to use this info to figure out our number of rose of atoms In a one m distance. So solving for that below, we would take our assumption being a one m side length. And we would divide by our height value which we related to the distance occupied by one row of atoms. And so we would divide by this value being .56 or meters. And so because we need to cancel our units to end up with rose as our final unit. We're going to multiply in a dimensional analysis step by our conversion factor where we recall that our previous PICO tells us we have 10 to 12 power PICO meters for one of our base unit meter. And so now we can cancel out meters as well as parking meters. And this leaves us with our implied unit, which is going to give us a value of 4.914 times 10 to the 9th. Power Rose. And again, because we can see above that, we have an offset square arrangement for arrangement too. We're going to calculate for the square. And so we would take our number of atoms that could be arranged in a one m distance, which above we agreed is equal to this value here. 4.256 times 10 to the ninth power. We're gonna plug that down below. And we want to multiply this by the number of rows that we can have in our one m length distance. And so we would multiply this by 4.914 times 10 to the ninth power for our rose. And so multiplying our atoms times the rows that should be present in our assumed one m length. We would get a value equal to 2.91 times to the 19th power atoms. And so this value here is also speaking to the number of atoms that can be packed in arrangement to or the density of arrangement to. And so now that we have this value, we want to compare this to get a ratio between our number of atoms for the area in arrangement one versus the area of arrangement to here. So we're going to divide them by each other. So putting the larger value in the numerator, we have 2.91 times 10 to the 19th power atoms. This was from arrangement two divided by our number of atoms and arrangement one which above we found to be 1. times 10 to the 19th power atoms in arrangement one. And so this gives us a ratio value equal to a value of 1.15. And so this is going to be The ratio where our arrangement to is 1.15 times that of arrangement one. So based on this value, we can say that therefore, arrangement too, Which is in two D. Has more atoms per square unit and shows an increase of potassium atoms arranged on the surface. That is 1.15 times that of arrangement one. And so for our final answer, we would say that we have a value of 1. for our ratio, which speaks to the higher density seen in arrangement too. So that would make sense based on our diagram because as we can see, even though our atoms are in an offset square, we have a smaller gap between our atoms versus in depiction one or arrangement one where we have a larger gap between our atoms. And so it definitely would make sense that we have a more denser arrangement in Arrangement to so 1.15 would be our final answer. I hope that everything that I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Verified Solution

Key Concepts

Atomic Arrangement

Close-Packing

Surface Density