Recent content by HelpPlease27

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    Single slit with minimum uncertainty

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

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

    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?
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    Single slit with minimum uncertainty

    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...
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    Compton Scattering Angle Calculation

    Thank you so much. I finally got it to work.
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    Compton Scattering Angle Calculation

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

    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
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    Compton Scattering Angle Calculation

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

    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)?
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    Compton Scattering Angle Calculation

    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 $$p=h/\lambda$$
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    Compton Scattering Angle Calculation

    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...
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    Relativity: Solving Pion Homework Problem

    Yes, that makes sense. I forgot to include gamma. I got the correct answer now, thanks.
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    Relativity: Solving Pion Homework Problem

    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
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