Question

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. $100keV$
• B. $500keV$
• C. $25keV$
• D. $1keV$

Answer 1

Answer: (C)

$25keV$

$\lambda =\frac{h}{p}\{\lambda =7.5\times {10}^{-12}\}$
$P=\frac{h}{\lambda }$
$KE=\frac{P^2}{2m}=\frac{(h/\lambda {)}^2}{2m}=\frac{\left\{\frac{6.6\times {10}^{-34}}{7.5\times {10}^{-12}}\right\}}{2\times 9.1\times {10}^{-31}}$
$KE=25KeV$.