Physics - Geometrical Optics Question with Solution | TestHub

In electron microscope, electrons are used as the 'light source'. The resolving power of a microscope is approximately equal to that of wavelength of the light used. The de Broglie wavelength $\lambda$ of a particle is related to its momentum $p$ by $$\lambda =\frac{h}{p}=\frac{h}{mv}$$ The speed of required electron is $$v=\frac{h}{m\lambda }=\frac{6.63\times 10^{-34}\mathrm{JS}}{9.11\times 10^{-31}\mathrm{kg}\cdot 10^{-11}\mathrm{m}}=7.28\times 10^7\mathrm{m}/\mathrm{s}$$ This corresponds to kinetic energy of $E=\frac{1}{2}m{v}^2$ $$\begin{array}{rl}& =\frac{1}{2}\times 9.11\times 10^{-31}\times {\left(7.28\times 10^7\right)}^2\times \frac{1\mathrm{eV}}{1.6\times 10^{-19}\mathrm{J}}\\ & =15.1\mathrm{keV}.\end{array}$$