One mole of tungsten has a mass of 184 grams, and its density is 19.3 grams per cubic centimeter, so the center-to-center distance between atoms is 2.51×10-10 m. You have a long thin bar of tungsten, 2.2 m long, with a square cross-section 0.16 cm on a side. You hang the rod vertically and attach a 418 kg mass to the bottom, and you observe that the bar becomes 0.983 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in tungsten. What is the spring stiffness of the entire wire, considered as a single macroscopic, very stiff spring?
How many side-by-side atomic chains are there in this wire? Assume that the tungsten atoms are arranged in a cubic lattice.
How many interatomic bonds are there in one atomic chain running the length of the wire?
What is the stiffness of a single interatomic "spring"?
Use the same approach I used for
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Well, well, well! Looks like we've got some serious scientific interrogation going on here! Don't worry, I've got my clown lab coat on and I'm ready to put some laughter into this tricky equation.
To find the spring stiffness of the entire wire, we need to determine the equivalent spring constant. But before we get there, let's break it down step by step.
First, let's calculate the change in length of the wire: 0.983 cm = 0.00983 m (converting to meters).
The force applied to the bottom of the wire is the weight of the 418 kg mass: F = m * g = 418 kg * 9.8 m/s^2 ≈ 4096.4 N.
Now, let's use Hooke's Law to find the spring constant (stiffness) of the entire wire. Hooke's Law states that F = k * ΔL, where F is the force applied, k is the spring constant, and ΔL is the change in length.
Rearranging the equation, we get k = F / ΔL ≈ 4096.4 N / 0.00983 m ≈ 416,764 N/m.
So, the spring constant of the whole wire is approximately 416,764 N/m.
Now, onto those side-by-side atomic chains! If we consider the square cross-section of the wire, it has an area of (0.16 cm)^2 = 0.0256 cm^2.
In one atomic chain, the number of tungsten atoms (side-by-side) is equal to the width of the wire divided by the center-to-center distance between atoms: 0.16 cm / (2.51×10^-10 m) = 6.37×10^8.
Then, multiply this by the length of the wire to get the total number of tungsten atoms in a single chain: 6.37×10^8 * 2.2 m = 1.4026×10^9.
Now, let's move onto the interatomic bonds. In a cubic lattice, each atom is connected to its six nearest neighbors.
Therefore, the number of interatomic bonds in one atomic chain can be calculated by multiplying the number of tungsten atoms in a chain by six: 1.4026×10^9 * 6 = 8.416×10^9.
And finally, the stiffness of a single interatomic "spring". To find this, we divide the spring constant of the whole wire by the number of interatomic bonds in one atomic chain: 416,764 N/m ÷ 8.416×10^9 ≈ 4.949×10^-8 N/m.
And there you have it! With a touch of clown magic and a whole lot of calculations, we've determined the stiffness of a single interatomic bond in tungsten. Now, go forth and spread the laughter of science!
To determine the spring stiffness of the entire wire, consider the following steps:
Step 1: Calculate the original length of the wire
The original length of the wire is given as 2.2 m.
Step 2: Calculate the change in length of the wire
The change in length of the wire is given as 0.983 cm, which is equivalent to 0.00983 m.
Step 3: Calculate the strain in the wire
The strain in the wire can be calculated using the formula:
strain = change in length / original length
strain = 0.00983 m / 2.2 m = 0.004469
Step 4: Determine the spring constant of the entire wire
The spring constant, or stiffness, of the wire can be calculated using Hooke's Law, which states that the force (F) applied to a spring is directly proportional to the extension or compression (x) of the spring.
F = kx
Rearranging the formula gives:
k = F / x
The force applied to the wire can be calculated using the mass of the attached load. The weight (force) of the load is given as 418 kg * 9.8 m/s^2 (acceleration due to gravity), which is equal to 4092.4 N.
Therefore, the spring constant of the entire wire is:
k = 4092.4 N / 0.00983 m = 416,676.42 N/m (or kg/s^2)
To determine the number of side-by-side atomic chains in the wire, consider the following steps:
Step 1: Calculate the cross-sectional area of the wire
The cross-sectional area of the wire can be calculated using the formula:
area = side length ^ 2
The side length of the square cross-section is given as 0.16 cm, which is equivalent to 0.0016 m.
Therefore, the cross-sectional area of the wire is:
area = (0.0016 m) ^ 2 = 2.56 × 10^-6 m^2
Step 2: Calculate the number of atomic chains in the wire
The number of atomic chains can be calculated by dividing the cross-sectional area by the area occupied by each atomic chain.
The area occupied by each atomic chain can be considered as a square with side length equal to the center-to-center distance between atoms, which is given as 2.51 × 10^-10 m.
Therefore, the number of atomic chains in the wire is:
number of chains = area / (side length of chain)^2
number of chains = 2.56 × 10^-6 m^2 / (2.51 × 10^-10 m)^2 = 407,571.71
To determine the number of interatomic bonds in one atomic chain running the length of the wire, consider the following steps:
Step 1: Calculate the number of atoms in one atomic chain
The number of atoms in one atomic chain can be calculated by dividing the length of the wire by the center-to-center distance between atoms.
The length of the wire is given as 2.2 m.
Therefore, the number of atoms in one atomic chain is:
number of atoms = length of wire / center-to-center distance
number of atoms = 2.2 m / (2.51 × 10^-10 m) = 8,764,940,240.64
Step 2: Calculate the number of interatomic bonds
The number of interatomic bonds in one atomic chain is equal to the number of atoms minus one.
Therefore, the number of interatomic bonds in one atomic chain is:
number of bonds = number of atoms - 1
number of bonds = 8,764,940,240.64 - 1 = 8,764,940,239.64
Finally, to determine the stiffness of a single interatomic "spring," consider dividing the spring constant of the entire wire by the number of interatomic bonds in one atomic chain.
Stiffness of a single interatomic "spring" = k / number of bonds
Stiffness of a single interatomic "spring" = 416,676.42 N/m / 8,764,940,239.64
Stiffness of a single interatomic "spring" ≈ 4.75 × 10^-11 N/m (or kg/s^2)
To determine the stiffness of one interatomic bond in tungsten, as well as the spring stiffness of the entire wire, we need to first solve for the number of side-by-side atomic chains in the wire, calculate the number of interatomic bonds in one atomic chain, and then use these values to find the stiffness of a single interatomic "spring."
1. Number of side-by-side atomic chains:
The cross-sectional area of the wire can be calculated by using the formula A = side^2, where side is the length of one side of the square cross-section (0.16 cm). Thus, A = (0.16 cm)^2 = 0.0256 cm^2.
The volume of the wire can be calculated by using the formula V = A * L, where L is the length of the wire (2.2 m) converted to cubic centimeters (1 m = 100 cm). Thus, V = 0.0256 cm^2 * (2.2 m * 100 cm/m)^3 = 5.632 cm^3.
Since tungsten has a density of 19.3 grams per cubic centimeter, we can calculate the mass of the wire using the formula mass = density * volume. Thus, mass = 19.3 g/cm^3 * 5.632 cm^3 = 108.6016 g.
Since one mole of tungsten has a mass of 184 grams, we can calculate the number of moles in the wire using the formula moles = mass / molar mass. Thus, moles = 108.6016 g / 184 g/mol = 0.590245 moles.
As per the question, the center-to-center distance between atoms is given as 2.51×10-10 m. We can now calculate the number of side-by-side atomic chains by dividing the length of the wire by the center-to-center distance. Thus, number of side-by-side atomic chains = (2.2 m * 100 cm/m) / (2.51×10-10 m) = 8.76494×10^11.
Therefore, there are approximately 8.76494×10^11 side-by-side atomic chains in the wire.
2. Number of interatomic bonds in one atomic chain:
Since the tungsten atoms are arranged in a cubic lattice, the number of interatomic bonds per atom is 12 (as each atom has 12 nearest neighbors, one in each direction).
Thus, the number of interatomic bonds per atomic chain running the length of the wire is 12.
3. Stiffness of a single interatomic "spring":
To calculate the stiffness of a single interatomic "spring," we need to know the change in length of the wire (0.983 cm) and the force acting on it (weight of the attached mass, which is equal to mass * acceleration due to gravity).
The force acting on the wire is given by the formula force = mass * acceleration due to gravity. Thus, force = 418 kg * 9.8 m/s^2 = 4096.4 N.
Using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position, we can calculate the stiffness of a single interatomic "spring" using the formula stiffness = force / change in length.
Thus, stiffness = 4096.4 N / 0.983 cm = 4168.78 N/cm.
Therefore, the stiffness of a single interatomic "spring" is approximately 4168.78 N/cm.