A) 15 keV
B) 1.5 keV
C) 150 keV
D) 1.5 MeV
Correct Answer: A
Solution :
The wavelength of light used in electron microscope is nearly equal to the resolving power of electron microscope. The de-Broglie wavelength is \[\lambda =\frac{h}{p}=\frac{h}{m\upsilon }\Rightarrow \upsilon =\frac{h}{m\lambda }\] ...(i) Here: \[\lambda \text{= }10\text{ }pm={{10}^{-11}}m,\] \[m=9.1\times {{10}^{-31}}kg\] and \[h=6.6\times {{10}^{-34}}Js\] So, from eq. (i), the speed of required electron. \[\upsilon =\frac{6.6\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times {{10}^{-11}}}=7.25\times {{10}^{-7}}m/s\] The energy of electron is\[=\frac{1}{2}m{{\upsilon }^{2}}\] \[=\frac{1}{2}\times 9.1\times {{10}^{-31}}\times {{(7.25\times {{10}^{7}})}^{2}}\times \frac{1\,eV}{1.6\times {{10}^{-19}}}J\] \[=15keV\]
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