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An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light ? |
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Answer» Solution :For an electron beam accelerated by a POTENTIAL V, the de-Broglie wavelength `lamda=(h)/(sqrt(2mVe))=(1.227)/(sqrt(V))=nm` In present case `V=50kV=50000V` `therefore lamda=(1.227)/(sqrt(50000))5.5xx10^(-3)nm` We know that resolving power of a microscope is GIVEN by `R.P.(2nsinalpha)/(1.22lamda)` [where `nsinalpha=`numeical APERTURE] As compared to yellow LIGHT `(lamda=550nm)`, the wavelength of electron wave `(lamda=5.5xx10^(-3)nm)` is `10^(-5)` times, hence resolving power of the electron microscope is about `10^(5)` times that of an optical microscope provided that numerical aperture is same. |
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