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The aperture of an electron microscope is 0.02, the accelerating potential is 10^(4) V Find the dimensions of details which may be resolved with the aid of this instrument.

Answer» <html><body><p><br/></p>Solution :We <a href="https://interviewquestions.tuteehub.com/tag/use-1441041" style="font-weight:bold;" target="_blank" title="Click to know more about USE">USE</a> the expression for the resolving power of a microscope (66.8), putting `sinu=0.02`. We can <a href="https://interviewquestions.tuteehub.com/tag/find-11616" style="font-weight:bold;" target="_blank" title="Click to know more about FIND">FIND</a> the wavelength from the nonrelativistic formula, since the kinetic <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of the electron of <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> keV is much <a href="https://interviewquestions.tuteehub.com/tag/less-1071906" style="font-weight:bold;" target="_blank" title="Click to know more about LESS">LESS</a> than its rest energy which is 510 keV.</body></html>


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