question_answer

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of \[7.5\times {{10}^{-}}^{12}\] m, the minimum electron required is close to- [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) 25 ke V                        

B) 500 ke V

C) 100 ke V                       

D)   1 ke V

Correct Answer: A

Solution :

\[\frac{d}{f}\,=\,\frac{1.22}{v}\,\,\frac{\lambda v}{a}\] \[d\,\,=\,\,1.22\,\,\frac{\lambda }{a}\] \[{{d}_{min}}\] is of order of wavelength of light \[=7.5\times {{10}^{-12}}=0.075\overset{{}^\circ }{\mathop{A}}\,\] \[=7.5\times {{10}^{-12}}=\frac{h}{\sqrt{2mKE}}\] \[0.075\,=\,\frac{12.26}{\sqrt{KE}}\] \[KE\,=\,{{\left( \frac{12.26}{0.075} \right)}^{2}}\,=\,\,25\,ke\,V\]