Answer:
Therefore the diameter of the orbit is 1.91m
Explanation:
The path followed by the particle is a circle that keeps the relation given by
[tex]BVq=\frac{mv^2}{r}[/tex]
Re-arrange the equation for r,
[tex]r=\frac{mv}{Bq}[/tex]
From V is equal to speed of ions and there is a procent for this 'real speed' we have
[tex]V=\frac{2.75}{100} 3*10^8m/s[/tex]
[tex]V= 8.25*10^6m/s[/tex]
Replacing in the previous equation,
[tex]r= \frac{(3.67*10^{-26})(8.25*10^6)}{(1.981)(1.6*10^{-19})}[/tex]
[tex]r=0.955m[/tex]
[tex]D=r*2= 1.91m[/tex]
Therefore the diameter of the orbit is 1.91m
