Question

An experiment was performed in which neutrons were shot through two slits spaced 0.10 mm apart and detected 3.5 m behind the slits. Figure P38.49 shows the detector output. Notice the $100\mathrm{\mu m}$scale on the figure. To one significant figure, what was the speed of the neutrons?

Short answer

The resultant value is$\mathrm{v}=197\times 8\frac{\mathrm{m}}{\mathrm{s}}$

Step-by-step solution

Given information

Neutrons were fired through two slits spaced 0.10 mm apart and detected 3.5 m behind the slits in an experiment.

Calculations

In a double slit experiment, we can start with an expression for fringe spacing.

$$\begin{gathered} \mathrm{\Delta y}=\frac{\mathrm{L\lambda }}{\mathrm{d}} \\ \mathrm{\lambda }=\frac{\mathrm{\Delta yd}}{\mathrm{L}}(\mathrm{express}\mathrm{\lambda }) \\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}\text{(de Broglie wavelength)} \\ \frac{\mathrm{\Delta yd}}{\mathrm{L}}=\frac{\mathrm{h}}{\mathrm{mv}}\left(\text{substitute}\mathrm{\lambda }\right) \\ \mathrm{v}=\frac{\mathrm{hL}}{\mathrm{m\Delta rd}}\left(\text{express}\mathrm{v}\right) \\ \mathrm{\Delta y}=70\mathrm{\mu m}\text{(we can see from graph in the text of the problem)} \\ \mathrm{v}=\frac{\left(6.63\times {10}^{-34}\right)\times \left(3\times 5\right)}{\left(1.67\times {10}^{-27}\right)\times \left(70\times {10}^{-6}\right)\times \left(0.1\times {10}^{-3}\right)}\text{(substitute)} \\ \mathrm{v}=197\times 8\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$