OUNT OF ENERGY LOST IN COMPTON SCATTERING EXPERIMENT

AI Thread Summary
In a Compton scattering experiment, X-rays scattered at 90 degrees experienced a 1% energy reduction, prompting a discussion on calculating the incident energy. Bob initially used an incorrect approach involving a quadratic equation for wavelength, leading to nonsensical results. A correction was suggested, emphasizing the relationship between initial and final energies using the photon energy equation. The recoil angle of the electron was also highlighted as being crucial for determining the wavelength change. Ultimately, accurate calculations based on these principles are essential for finding the incident energy of the X-rays.
Bobster
Messages
16
Reaction score
0
Any ideas how to do this question ?

In a Compton type experiment X-rays scattered through 90 degrees suffered one percent reduction in energy.What was the incident energy of the X-rays ? You can assume that the electron coiled non-relativistically.

Any help would be really appreciated ! :smile:

Bob
 
Physics news on Phys.org
Oh by the way,i used the equation :


dL=h/mc (1-cos angle)

and i got dL=2.43 E -12

Since the energy reduction is 1 % i did

hc/L-hc/L' =0.01

and L'=L+2.43 E -12

so we end up with a quadratic :

L^2 - (2.43 E -12)L-(4.83 E -35)=0

Solving I got, L= O or -2.43 E -12

Hence incident energy is hc/0 or hc/-2.43 E -12

=> energy =infinity or hc/(-2.43 E -12). :confused:

Now is this total rubbish,or am I getting somewhere ? :bugeye:

Cheers for any help !
Bob
 
I believe you made an error when calculating the final energy (and hence wavelength) of the X-ray. Do it like this. The final energy of the X-ray E_f = 0.99E_i where the i stands for "initial". Now use the Einstein relation for the energy of a photon: E = \frac{hc}{\lambda} to figure out the relationship between the final and initial wavelengths of the X-rays. You shouldn't get any type of quadratic term in \lambda
 
Thanks mate ! that was exactly my mistake.I can't believe I made that error :cry: but oh well !. :rolleyes:

Cheers again !

This world needs more people like :approve: you
 
Bobster said:
Any ideas how to do this question ?

In a Compton type experiment X-rays scattered through 90 degrees suffered one percent reduction in energy.What was the incident energy of the X-rays ? You can assume that the electron coiled non-relativistically.

Any help would be really appreciated ! :smile:

Bob
I think the key is to get the recoil angle of the electron. It is almost 45 degrees (tan\theta = .99). Plug that into the Compton formula and you will get the magnitude of the wavelength change. That change represents a loss of one percent of the original energy.

AM
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top