Who's up for a challenge? Quantum Que

[minimax]

Homework Statement

An incident x-ray has a wavelength of 0.525nm and scatters as 112 degrees above the incidental x-axis in the positive x-direction. Find the direction of propagation of the electron that the incidental wavelength hits. (the answer is to be 34 degrees below the x-axis)

Homework Equations

Using Compton's Effect
$$\frac{h}{\lambda_{1}}$$=$$\frac{h}{\lambda_{2}}$$ cos$$\theta$$+pecos$$\phi$$

0=$$\frac{h}{\lambda_{2}}$$ sin$$\theta$$+pesin$$\phi$$

where $$\phi$$ = angle of electron direction
$$\theta$$= angle of x-ray direction
pe=momentum of electron

The Attempt at a Solution

Since the electron momentum is not given, numbers can't be plugged into find the angle.
Possibly the two equations have to be manipulated together (ie add/subtract) somehow?
Or using energy equations?

 

[anathema]

Thank you for your question. I believe that the correct approach to solving this problem would be to use the Compton scattering equation, which is given by:

\frac{h}{\lambda_{1}}=\frac{h}{\lambda_{2}} cos\theta+pecos\phi

where h is the Planck's constant, \lambda_{1} is the incident x-ray wavelength, \lambda_{2} is the scattered x-ray wavelength, \theta is the angle of the incident x-ray in relation to the x-axis, pe is the momentum of the electron, and \phi is the angle of the electron direction.

In order to solve for the angle of the electron direction, we can rearrange the equation to isolate \phi:

\phi = cos^{-1} \left( \frac{\frac{h}{\lambda_{1}} - \frac{h}{\lambda_{2}} cos\theta}{pe} \right)

We can then plug in the given values:

\phi = cos^{-1} \left( \frac{\frac{6.626 \times 10^{-34} J s}{0.525 \times 10^{-9} m} - \frac{6.626 \times 10^{-34} J s}{\lambda_{2}} cos112^\circ}{pe} \right)

Since we do not have the momentum of the electron, we can use the fact that energy is conserved in Compton scattering to find the scattered x-ray wavelength, which is related to the energy of the electron. The energy of the incident x-ray can be calculated using the equation:

E = \frac{hc}{\lambda_{1}}

where c is the speed of light. We can then use the conservation of energy equation:

E_{1} + E_{2} = E_{3}

where E_{1} is the energy of the incident x-ray, E_{2} is the energy of the electron, and E_{3} is the energy of the scattered x-ray. We can then rearrange the equation to solve for the energy of the electron:

E_{2} = E_{3} - E_{1}

Since we know the energy of the incident x-ray and we can calculate the energy of the scattered x-ray using the scattered x-ray wavelength, we can then solve for the energy of the electron. Once we have the energy of the electron, we can

 

[Moetasim]

I am always up for a challenge! Let's start by breaking down the problem and identifying the relevant equations. We are given an incident x-ray with a specific wavelength and a scattering angle. We are asked to find the direction of propagation of the electron that the incident x-ray hits. This scenario can be described using Compton's Effect, which relates the change in wavelength of a scattered photon to the momentum of the electron it interacts with.

To solve this problem, we will use the following equations:

1. \frac{h}{\lambda_{1}}=\frac{h}{\lambda_{2}} cos\theta+pecos\phi

2. 0=\frac{h}{\lambda_{2}} sin\theta+pesin\phi

where \phi is the angle of the electron's direction, \theta is the angle of the x-ray direction, and pe is the momentum of the electron.

To find the angle of the electron's direction, we need to solve for \phi in these equations. We can do this by manipulating the equations to eliminate pe. To do this, we can square both equations and add them together. This will cancel out the term involving pe, and we will be left with an equation that only involves \phi.

After solving for \phi, we can plug in the given values for \theta and solve for the angle of the electron's direction. In this case, the angle is found to be 34 degrees below the x-axis.

Alternatively, we can also use energy equations to solve this problem. According to the law of conservation of energy, the energy of the incident x-ray must equal the sum of the energy of the scattered x-ray and the energy of the electron. By equating these energies, we can solve for the angle of the electron's direction.

In conclusion, by using Compton's Effect and the law of conservation of energy, we can determine the angle of the electron's direction in this scenario. This type of problem is common in quantum mechanics and requires a solid understanding of fundamental principles and equations. I hope this explanation helps in understanding the solution to this challenge.