KVPY Sample Paper KVPY Stream-SX Model Paper-24

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 energy required is close to:

A) 500 ke V

B) 100 ke V

C) 1 ke V

D) 25 ke V

Correct Answer: D

Solution :

\[k=\frac{{{P}^{2}}}{2m}=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\]

\[=\frac{{{(6.62\times {{10}^{-34}})}^{2}}}{2\times 9.1\times {{10}^{-31}}\times (7.5\times {{10}^{-12}})}\times \frac{1}{1.6\times {{10}^{-19}}}eV\]

Alternate method\[\lambda =\sqrt{\frac{150}{V}}\] \[\Rightarrow \] \[7.5\times {{10}^{-12}}=\sqrt{\frac{150}{V}}\]

\[V=\frac{80}{3}kV\]

Energy \[=\frac{80}{3}keV\simeq 25keV\]