I tried the equation E = hc/ = (6.63e-34)(3e8)/(1e-12) =1.24e6 eV (b) If a light

Question

I tried the equation E = hc/ = (6.63e-34)(3e8)/(1e-12) =1.24e6 eV

(b) If a light microscope is used, what minimum photon energy isrequired?
keV

(c) Which microscope seems more practical? Why? Thelight microscope, since it needs less energy.Thelight microscope, since it needs moreenergy.    Theelectron microscope, since it needs lessenergy.Theelectron microscope, since it needs moreenergy.Bothare equally practical, since they require about the sameenergy. Thelight microscope, since it needs less energy.Thelight microscope, since it needs moreenergy.    Theelectron microscope, since it needs lessenergy.Theelectron microscope, since it needs moreenergy.Bothare equally practical, since they require about the sameenergy. Thelight microscope, since it needs less energy.Thelight microscope, since it needs moreenergy.    Theelectron microscope, since it needs lessenergy.Theelectron microscope, since it needs moreenergy.Bothare equally practical, since they require about the sameenergy.

Explanation / Answer

E=hf this equation is only for the light.( you can findthis in the book. Einstein said Energy of the light is E^2 = (pc)^2 + (mc^2)^2 but the mass of the photonis 0 . so E=hf . this means E=hf is only for light.) this question ask enegy for electron. so you can use the equation E=sqrt(p^2/2m) ,p=h/lambda go a head~

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