# To resolve an object in an electron microscope, the wavelength of the electrons must be close to the diameter of the object. What kinetic energy must the electrons have in order to resolve a protein molecule that is 2.40 nm in diameter?

## Take the mass of an electron to be $9.11 x {10}^{-} 31$ kg.

May 5, 2017

$\textsf{4.18 \times {10}^{- 20} \textcolor{w h i t e}{x} J}$

#### Explanation:

I will set the wavelength of the electron to be equal to the diameter of the protein molecule:

$\textsf{\lambda = 2.4 \times {10}^{- 9} \textcolor{w h i t e}{x} m}$

The de Broglie expression gives us:

$\textsf{\lambda = \frac{h}{m v}}$

$\therefore$$\textsf{v = \frac{h}{m \lambda}}$

$\textsf{v = \frac{6.63 \times {10}^{- 34}}{9.11 \times {10}^{- 31} \times 2.4 \times {10}^{- 9}} \textcolor{w h i t e}{x} \text{m/s}}$

$\textsf{v = 3.03 \times {10}^{5} \textcolor{w h i t e}{x} \text{m/s}}$

Kinetic energy $\textsf{= \frac{1}{2} m {v}^{2}}$

$\therefore$$\textsf{K E = \frac{1}{2} \times 9.11 \times {10}^{- 31} \times {\left(3.03 \times {10}^{5}\right)}^{2} \textcolor{w h i t e}{x} J}$

$\textsf{K E = 4.18 \times {10}^{- 20} \textcolor{w h i t e}{x} J}$