Recent content by HelpPlease27

Ok, I get it. Thank you

So then is $$\tan\theta * x = y$$ which is the central diffraction maximum? Or am I still missing something?

Thanks for explaining it

I get this part but not the next part. I still don't get what the spread in the y-direction corresponds to?

I don't understand how I can use that to work out the central diffraction maximum. $$\Delta Py*\Delta y<h/4\pi$$ So is $$\Delta y$$ the central diffraction maximum? Where does the x component come into it?

Homework Statement A horizontal beam of laser light of wavelength 474 nm passes through a narrow slit that has width 5.80×10−2 mm . The intensity of the light is measured on a vertical screen that is 2.40 m from the slit. Use the result of part A to estimate the width of the central diffraction...

Thank you so much. I finally got it to work.

I'm still not getting it to work. I'm now getting $$\lambda' = 1.715*10^-12$$

I did it for both and the difference is small so I guess either works. I got $$\lambda = 7.79*10^-11$$ and when I used this in the equation I can't get it to work. It worked out to be $$\cos\phi = -23.99$$, which doesn't work

Yea, it would be so for energy I need to use K = mc^2/sqrt(1-v^2/c^2) - mc^2 ?

Can I use that the kinetic energy is 1/2mv^2 with m = mass of an electron(?) and the v = 9.3 * 10 ^6 m/s. Then the wavelength is just h/sqrt(2*E*m)?

I assumed it meant the scattered wavelength but is it the original wavelength and I need to find the scattered wavelength? I can use it to get the wavelength I need because for the photons $$p=h/\lambda$$

Homework Statement A photon with wavelength 0.1385 nm scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is 9.30×106 m/s? Homework...

Yes, that makes sense. I forgot to include gamma. I got the correct answer now, thanks.

E^2 = (mc^2)^2 + (pc)^2 I used mc^2 = 139.6 MeV I put p = mv so the pc = mvc but m = 139.6/c^2 and v = (1-Δ)c = (1-(3.04*10-5))c so pc = 139.6 MeV So then E = sqrt(139.6^2 + 139.6^2) = 197.4 MeV