dual-nature-of-radiation-and-matter Question 46
Question: Q. 8. An electron microscope uses electrons accelerated by a voltage of $50 \mathrm{kV}$. 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]
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Solution:
Ans. } \begin{array}{rlr} \lambda & =\frac{h}{p}=\frac{h}{\sqrt{2 m e V}} \text { or } \lambda=\frac{12.27}{\sqrt{V}} \AA & 1 / 2 \ \therefore \quad \lambda & =\frac{6.63 \times 10^{-34}}{\sqrt{\left(2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 50 \times 10^{3}\right)}} & 1 / 2 \ \lambda & =5.33 \times 10^{-12} \mathrm{~m} & 1 / 2 \end{array} $$
The resolving power of an electron microscope is much better than that of optical microscope. $1 / 2$
[CBSE Marking Scheme 2014]
Detailed Answer :
De-Broglie wavelength of an electron
Accelerated pôtential $=50 \times 10^{3} \mathrm{~V}$
Hence, $O$
$$ \begin{aligned} \lambda_{e} & =\frac{1.227}{\sqrt{50 \times 10^{3}}} \ & =0.548 \times 10^{-2} \mathrm{~nm} \ & =5.48 \times 10^{-12} \mathrm{~m} \end{aligned} $$
Resolving Power of Microscope
$$ P=\frac{2 n \sin \beta}{1.22 \lambda} $$
From the formula, it is clear that resolving power increases as wavelength decreases keeping other factors as constant.
Wavelength of yellow light $=680 \mathrm{~nm}=6.8 \times 10^{-7} \mathrm{~m}$ 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.
