Question

Electrons, all with the same speed, pass through a tiny 15-nm-wide slit and create a diffraction pattern on a detector 50 mm behind the slit. What is the electron’s kinetic energy, in eV, if the central maximum has a width of 3.3 mm?

Short answer

The electron’s kinetic energy is${\mathrm{E}}_{\mathrm{k}}=2507\mathrm{eV}$

Step-by-step solution

Given information

The diffraction pattern on a detector 50 mm behind the slit .

Making diagram of diffraction pattern 

Calculations

Starting with a diffraction expression on one slit, we can solve the problem.

$$\begin{gathered} \mathrm{dsin}{\mathrm{\theta }}_{\mathrm{n}}=\mathrm{n\lambda } \\ \mathrm{dsin}\mathrm{\theta }=\mathrm{\lambda }(\mathrm{n}=1\text{central maximum}) \\ \tan \mathrm{\theta }=\frac{\mathrm{\Delta y}}{2\mathrm{L}}\left(\text{from picture}\right) \\ \mathrm{\theta }=\arctan \frac{\mathrm{\Delta y}}{2\mathrm{L}}\left(\text{express}\mathrm{\theta }\right) \\ \mathrm{\lambda }=\mathrm{dsin}\left(\arctan \left(\frac{\mathrm{\Delta y}}{2\mathrm{L}}\right)\right)\left(\text{substitute}\mathrm{\theta }\right) \\ {\mathrm{E}}_{\mathrm{k}}=\frac{\mathrm{hc}}{\mathrm{\lambda }}\text{(kinetic energy)} \\ {\mathrm{E}}_{\mathrm{k}}=\frac{\mathrm{hc}}{\mathrm{dsin}\left(\arctan \left(\frac{\mathrm{\Delta y}}{2\mathrm{L}}\right)\right)}\text{(substitute}\mathrm{\lambda }\text{)} \\ {\mathrm{E}}_{\mathrm{k}}=\frac{\left(6.63\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{15\cdot {10}^{-9}\cdot \sin \left(\arctan \left(\frac{3.3\times {10}^{-3}}{2\cdot 50\times {10}^{-3}}\right)\right)}\text{(substitute)} \\ {\mathrm{E}}_{\mathrm{k}}=2507\mathrm{eV} \end{gathered}$$