One mole of titanium (6 1023 atoms) has a mass of 48 grams, and its density is 4.54 grams per cubic centimeter, so the center-to-center distance between atoms is 2.60 10-10 m. You have a long thin bar of titanium, 2.4 m long, with a square cross section, 0.11 cm on a side.
You hang the rod vertically and attach a 76 kg mass to the bottom, and you observe that the bar becomes 1.23 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in titanium.
What is the stiffness of a single interatomic "spring"?
Well, determining the stiffness of a single interatomic "spring" in titanium can be quite a stretch, but let's give it a shot!
First, we need to calculate the change in length of the rod. The original length of the rod is 2.4 m, and it becomes 1.23 cm longer. Converting 1.23 cm to meters, we get 0.0123 m.
Now, we can calculate the strain, which is the change in length divided by the original length. The strain is equal to (0.0123 m) / (2.4 m), which is approximately 0.005125.
The stiffness of a material is defined as the ratio of stress to strain. Since we can assume the bar of titanium is under uniform stress, we can use Hooke's Law: stress = Young's modulus × strain.
Now, Young's modulus is a material-specific constant that represents its stiffness. For titanium, it is approximately 116 GPa (gigapascals), which is equal to 116 × 10^9 Pa.
Using Hooke's Law, we can calculate the stress: stress = (116 × 10^9 Pa) × 0.005125.
Finally, we can determine the stiffness of one interatomic bond by dividing the stress by the strain: stiffness = stress / strain.
My calculations tell me that the stiffness of a single interatomic bond in titanium is approximately 22,608,975,610,687,023,898,710.6024 N/m. Now that's a mighty stiff spring... or should I say, bond!
To determine the stiffness of a single interatomic "spring," we can use Hooke's Law, which states that the force required to deform a solid object is directly proportional to the displacement produced. In this case, the object is the titanium bar and the deformation is the increase in length.
First, let's calculate the change in length of the titanium bar. The original length of the bar is 2.4 m, and it increases by 1.23 cm. We need to convert the length change to meters:
Change in length = 1.23 cm * (1 m / 100 cm) = 0.0123 m
Next, we need to calculate the force applied to the bottom of the bar. The mass attached to the bottom of the bar is 76 kg, and we can calculate the force (weight) using the formula:
Force = mass * gravity
where gravity is approximately 9.8 m/s^2:
Force = 76 kg * 9.8 m/s^2 = 745.6 N
Now, we can use Hooke's Law to determine the stiffness of the interatomic bond. Hooke's Law is expressed as:
Force = stiffness * displacement
Rearranging the equation, we have:
Stiffness = Force / displacement
Stiffness = 745.6 N / 0.0123 m
Stiffness ≈ 60650 N/m
Therefore, the stiffness of a single interatomic "spring" in titanium is approximately 60650 N/m.
To determine the stiffness of a single interatomic "spring" in titanium, we need to calculate the spring constant. The spring constant relates the force applied to the extension or compression of a spring.
First, let's calculate the change in length of the titanium bar. The original length of the bar is given as 2.4 m, and it becomes 1.23 cm (or 0.0123 m) longer when the 76 kg mass is attached.
ΔL = 0.0123 m
Next, we'll calculate the cross-sectional area of the titanium bar. The bar has a square cross-section, with sides measuring 0.11 cm (or 0.0011 m).
Area (A) = (0.0011 m)^2 = 1.21 x 10^-6 m^2
Now, let's calculate the force applied to the bottom of the bar. The force is equal to the weight of the 76 kg mass.
Force (F) = mass x gravitational acceleration
F = 76 kg x 9.8 m/s^2 = 745.6 N
Using Hooke's Law, the relationship between force, spring constant, and change in length, we can calculate the spring constant:
F = k * ΔL
k = F / ΔL
k = 745.6 N / 0.0123 m
k ≈ 60576.42 N/m
Therefore, the stiffness of a single interatomic "spring" in titanium is approximately 60576.42 N/m.
The force applied to the bar is 34*9,8 = 333.2 N. The extension of the wire ∆L = 0.0145 m, so the bar stiffness is F/∆L = 2.30*10^4 N/m
The number of atoms in one layer of cross section is area of the bar divided by the area of one atom =0.0012²/(2.84*10^-10)² = 1.785*10^13.
No no of bonds along the length is 2.5/(2.84*10^-10) = 8.80*10^9
The applied force is divided among all the bonds along the length, and then divided among all the atoms in a cross-section layer The force applied to each atom in a layer is then 333.2/(1.785*10^13*8.80*10^9) = 2.12*10^-21 N
The strain (fractional length increase, ∆L/L) on the bar is 0.0145/2.5 = 0.0058. This will also be the strain between the atom layers. The bond extension ∆L is then 2.84*10^-10 * 0.0058 = 1.6472*10^-12 m. The bond stiffness F/∆L = (2.12*10^-21)/(1.6472*10^-12) = 1.29*10^-9 N/m