Question

You are asked to consult for the city’s research hospital, where a group of doctors is investigating the bombardment of cancer tumors with high-energy ions. As figure shows, ions are fired directly toward the center of the tumor at speeds of $5.0\times {10}^6\mathrm{m}/\mathrm{s}$. To cover the entire tumor area, the ions are deflected sideways by passing them between two charged metal plates that accelerate the ions perpendicular to the direction of their initial motion. The acceleration region is $5\mathrm{cm}$long, and the ends of the acceleration plates are $1.5\mathrm{m}$from the target. What sideways acceleration is required to deflect an ion $2\mathrm{cm}$to one side?

Short answer

$\mathrm{a}=6.8\times {10}^{12}\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Given information

Speed is $\mathrm{v}=5.0\times {10}^6\mathrm{m}/\mathrm{s}$

The acceleration region is ${\mathrm{L}}_1=5.0\mathrm{cm}$long

The ends of the acceleration plates are ${\mathrm{L}}_2=1.5\mathrm{m}$from the target.

We have the determine the sideways acceleration to deflect an ion $\mathrm{d}=2.0\mathrm{cm}$to one side?

Calculation

Putting values:

The sideways acceleration is:

$$\begin{gathered} \mathrm{a}=\frac{2{\mathrm{v}}^2\mathrm{d}}{2{\mathrm{L}}_1{\mathrm{L}}_2+{{\mathrm{L}}_1}^2} \\ \mathrm{d}=2\mathrm{cm}=0.02\mathrm{m} \\ {\mathrm{L}}_1=5\mathrm{cm}=0.05\mathrm{m} \\ \mathrm{a}=\frac{2\times {(5\times {10}^6)}^2\mathrm{m}/\mathrm{s}\times 0.02\mathrm{m}}{2\times 0.05\mathrm{m}\times 1.5\mathrm{m}+{(0.05\mathrm{m})}^2} \\ \mathrm{a}=\frac{1\times {10}^{12}{\mathrm{m}}^2/\mathrm{s}}{0.15{\mathrm{m}}^2+2.5\times {10}^{-3}{\mathrm{m}}^2} \\ \mathrm{a}=\frac{2\times {10}^5{\mathrm{m}}^2/\mathrm{s}}{0.1525{\mathrm{m}}^2} \\ \mathrm{a}=6.56\times {10}^{12}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$