Tennis ball diffraction angles

[natugnaro]

Homework Statement

Quantum effects are negligible in "macroscopic" world.
Show than on example.
A tennis ball with speed 0.5 m/s is thrown thru a window.
Dimensions of a window are 1*1.5 m.
Calculate vertical and horizontal diffraction angles.

Homework Equations

[tex]\lambda[/tex]=h/p
dSin[tex]\alpha[/tex]=n[tex]\lambda[/tex]

The Attempt at a Solution

For horizontal diffraction angle.
[tex]\lambda[/tex]=h/mv=h/m*0.5

I assumed that n=1.
so

dSin[tex]\alpha[/tex]=n[tex]\lambda[/tex]=h/m*0.5

[tex]\alpha[/tex]=ArcSin(h/m*0.5*d)=ArcSin(h/m*0.5*1)

from this I would conclude thath, alpha is negligible small, that is
quantum effect are negligible.
I would do the same thing for vertical angle.
I' not shure if I'm right ?

 

[anathema]

I would like to clarify that the statement "quantum effects are negligible in the macroscopic world" is not entirely accurate. While it is true that quantum effects are usually not observable in macroscopic objects, they still play a crucial role in determining the behavior of these objects.

In the case of a tennis ball being thrown through a window, quantum effects may not have a significant impact on the trajectory of the ball. However, they do play a role in determining the properties of the ball itself, such as its mass and momentum.

To calculate the diffraction angles in this scenario, we can use the equations provided. The de Broglie wavelength of the tennis ball can be calculated using the equation \lambda = h/p, where h is Planck's constant and p is the momentum of the ball. Assuming the mass of the tennis ball is 0.057 kg (typical for a tennis ball), and its velocity is 0.5 m/s, we can calculate the wavelength to be approximately 1.22 x 10^-34 meters.

Next, we can use the equation dSin\alpha=n\lambda to calculate the diffraction angle. In this case, the value of n is not necessarily 1, as it depends on the number of diffraction patterns observed. However, for simplicity, we can assume n=1.

For the horizontal diffraction angle, we can use the width of the window (1.5 m) as the distance between the two slits, and the wavelength we calculated earlier to be the value of d. Plugging these values into the equation, we get Sin\alpha = (1)(1.22 x 10^-34)/(1.5) = 8.13 x 10^-35. Taking the inverse sine of this value, we get a horizontal diffraction angle of approximately 8.13 x 10^-35 radians.

Similarly, for the vertical diffraction angle, we can use the height of the window (1 m) as the distance between the two slits, and the same wavelength we calculated earlier to be the value of d. Plugging these values into the equation, we get Sin\alpha = (1)(1.22 x 10^-34)/(1) = 1.22 x 10^-34. Taking the inverse sine of this value, we get a vertical diffraction angle of approximately 1.22 x 10^-34 radians.

As

 

[ogb p]

Your solution is on the right track, but there are a few things to consider. First, the given dimensions of the window (1*1.5 m) do not provide enough information to calculate the diffraction angles. We would also need to know the size and shape of the tennis ball, as well as the distance from the window at which it is thrown. Without this information, it is not possible to accurately calculate the diffraction angles.

However, assuming the tennis ball is a standard size and shape and is thrown from a distance that is much larger than the dimensions of the window, we can make some estimations. In this case, the diffraction angles would indeed be negligible, as the size of the tennis ball is much larger than the quantum scale. This means that the diffraction pattern would be minimal and not noticeable to the naked eye.

In the macroscopic world, quantum effects are negligible because the objects and distances involved are much larger than the quantum scale. This means that the diffraction angles in this scenario would be very small and would not have a significant impact on the trajectory of the tennis ball. However, in the microscopic world, quantum effects are significant and must be taken into account in order to accurately describe and predict the behavior of particles.