can someone help me with this The image magnification that can be obtained with

Question

can someone help me with this

The image magnification that can be obtained with an optical microscope is limited by diffraction; it cannot resolve features of the sample that are smaller than the wavelength of the light, more - or - less. There are a few clever ways to push past this limit to some degree, but the relatively "large" wavelength of light still sets the scale for what image resolution can be obtained. On the other hand, an electron microscope (diagram at right is from Wikipedia) can produce stunning images of tiny things because the electron de Broglie wavelength can be made much smaller than the wavelength of visible light. What kinetic energy would an electron have to have to have a wavelength equal to that of blue light? What is the wavelength of an electron that has a kinetic energy of 10.0 keV? If the electron beam is adjusted to have a wavelength of 0.2 nm, what is the speed of the electrons?

Explanation / Answer

a) lamda = h/Sqrt(2mE) => E = (h/lamda)^2/2m = (6.626*10^-34/4000*10^-10)^2/(2*9.1*10^-31) = 9.42*10^-9 eV

b) lamda = h/(Sqrt(2mE)) = (6.626*10^-34/(sqrt(2*9.1*10^-31*10*1.6*10^-19)) = 0.012 nm

c)lamda = h/mv => v = (6.626*10^-34/(9.1*10^-31*0.2*10^-9)) = 3.6*10^6 m/s

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