You can't get any help from anyone here if you say this. This is just ##\textbf{put-into-the-formula}## problemI don't really know how to proceed. Can someone help me out in this?
2moc^2, my bad.PhysicsTruth said:Yeah, but on which formula should I use it? I'm confused. For delta lambda to be minimum, theta had to be 0. But then, how is the energy moc^2? Really, I don't get it.
phyzguy said:Do you know how to use calculus to maximize a function?
From the problem statement, it should be apparent that the relevant function is the energy of the scattered photon as a function of angle. Call the energy of the incoming photon ##E_0##, and call the energy of the scattered photon ##E##. Can you come up with a function ##E(\theta)##?PhysicsTruth said:Yeah, but on which formula should I use it?
I get: ##1/E##= ##1/E_0## + (1-cos(theta))/moc^2jtbell said:From the problem statement, it should be apparent that the relevant function is the energy of the scattered photon as a function of angle. Call the energy of the incoming photon ##E_0##, and call the energy of the scattered photon ##E##. Can you come up with a function ##E(\theta)##?
Hint: what is the relationship between ##E## and ##\lambda##?
Can you please guide me how should I proceed further?PhysicsTruth said:I get: ##1/E##= ##1/E_0## + (1-cos(theta))/moc^2
But for E to be max, theta should be 0,so how to do it now?
Indeed! and what does E equal in that case?PhysicsTruth said:But for E to be max, theta should be 0
The Compton Effect is a phenomenon in which a photon (a packet of energy) collides with an electron, causing the photon to lose some of its energy and the electron to gain some of that energy. This results in a change in the wavelength and direction of the scattered photon.
The Compton Effect is a key piece of evidence for the wave-particle duality of light, as it shows that light can behave as both a wave and a particle. The scattered photon behaves like a particle, transferring its energy to the electron, while the change in wavelength demonstrates the wave-like nature of light.
The energy of a scattered photon is directly proportional to the angle of scattering. This means that the greater the angle of scattering, the more energy the photon loses in the collision with the electron.
The energy of a scattered photon can be calculated using the Compton formula: Escattered = Einitial / (1 + (Einitial/melectronc2)(1 - cosθ)), where Einitial is the initial energy of the photon, melectron is the mass of the electron, c is the speed of light, and θ is the angle of scattering.
The Compton Effect has many practical applications, including in medical imaging techniques such as X-rays and PET scans. It is also used in materials science to study the structure of materials, and in astrophysics to measure the energy of photons emitted by celestial objects.