Question

The electrons in a cathode-ray tube are accelerated through a $250\mathrm{V}$potential difference and then shot through a 33-nm-diameter circular aperture. What is the diameter of the bright spot on an electron detector $1.5\mathrm{m}$behind the aperture?

Short answer

The short answer of this sum is

$\Delta y=8.6·{10}^{-3}\mathrm{m}=8.6\mathrm{mm}$

Step-by-step solution

Step: 1 Given information

The diameter of the bright spot on an electron detector$1.5\mathrm{m}$

Figure shown below

Calculation

We can begin by focusing on energy conservation.

$$\begin{gathered} E_k=U \\ \frac{m{v}^2}{2}=e\Delta V \\ v=\sqrt{\frac{2e\Delta V}{m}} \\ \lambda =\frac{h}{mv} \\ \lambda =\frac{h}{m\sqrt{\frac{2e\Delta V}{m}}} \\ \sin \theta =1.22\frac{\lambda }{D} \\ \theta =\arcsin 1.22\frac{\lambda }{D} \\ \theta =\arcsin 1.22\frac{\frac{h}{m\sqrt{\frac{2·\Delta V}{m}}}}{D} \end{gathered}$$

From a picture and simple geometry, we can write.

$$\begin{gathered} \Delta y=2L\tan \theta \\ \Delta y=2L\mathrm{tanarcsin}1.22\frac{\frac{h}{m\sqrt{\frac{2e\Delta V}{m}}}}{D} \\ \Delta y=2·1.5·\mathrm{tanarcsin}1.22\frac{\frac{6.63·{10}^{-34}}{9.11·{10}^{-31·}\sqrt{\frac{2·1.6-{10}^{-19·250}}{9.11·{10}^{-31}}}}}{33·{10}^{-9}} \\ \Delta y=8.6·{10}^{-3}\mathrm{m}=8.6\mathrm{mm} \end{gathered}$$