How does energy and speed affect sub-atomic particles (protons, neutrons, electrons) in an electric field [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 4 years ago.








[Improve this question](/posts/114893/edit) 



[![enter image description here](https://i.stack.imgur.com/CxvbC.png)](https://i.stack.imgur.com/CxvbC.png)


(Source: <https://gyazo.com/a4d5beed3272a7ca793dd8e2f7e8be3c>)


I am having trouble understanding the problem above. Why does it matter if the particles are held at constant energy or speed? How does energy and speed even affect sub-atomic particles. Why do the different constants produce different diagrams?




The energy and speed do not affect particles themselves, but their interactions with other diatomic objects. 


And, of course, they affect their kinematics in electromagnetic field.


Kinetic energy of a particle is 


$$E\_\mathrm{k}=\frac 12 \cdot m \cdot v^2$$


The force on a charged particle in an electrostatic field of the strength $E$ is 


$$\vec F=q\cdot \vec E$$


The perpendicular acceleration is 


$$\vec a=\vec F/m=\frac{q\cdot \vec E}{m}$$


For the length of the path between the electrodes $L$, the time of flight is


$$t=\frac Lv=L \cdot \sqrt{ \left( \frac {m}{2\cdot E\_\mathrm{k}}\right)}$$


The perpendicular deviation due the acceleration in the field is 


$$\begin{align}
\vec L\_\mathrm{p}&=\frac 12 \cdot a \cdot t^2 \\
&= \frac 12 \cdot \frac{q\cdot \vec E}{m} \cdot \left( \frac {L^2\cdot m}{2\cdot E\_\mathrm{k}}\right) \\
&= \frac 14 \cdot q \cdot L^2 \cdot \frac{\vec E}{E\_\mathrm{k}}\\
\end{align}$$


One can see the **bending with the equal kinetic energy is mass independent.**


The scenario with the **equal speed** is a different story, as the **perpendicular acceleration has the reciprocal proportionality to the mass**, so the path of the much lighter electron is bent much more.


 1