Question: Q. 8. An electron microscope uses electrons accelerated by a voltage of 50kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

A [O.D. I, II, III 2014]

$$ \text {

Show Answer

Solution:

Ans. } λ=hp=h2meV or λ=12.27V\AA1/2 λ=6.63×1034(2×9.1×1031×1.6×1019×50×103)1/2 λ=5.33×1012 m1/2 $$

The resolving power of an electron microscope is much better than that of optical microscope. 1/2

[CBSE Marking Scheme 2014]

Detailed Answer :

De-Broglie wavelength of an electron

Accelerated pôtential =50×103 V

Hence, O

λe=1.22750×103 =0.548×102 nm =5.48×1012 m

Resolving Power of Microscope

P=2nsinβ1.22λ

From the formula, it is clear that resolving power increases as wavelength decreases keeping other factors as constant.

Wavelength of yellow light =680 nm=6.8×107 m As we have seen from numerical calculation that electron wavelength is much lower than yellow light, hence resolving power of electronic microscope is much better than optical microscope.



विषयसूची

जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक