Answer:
[tex]0.025\ \text{m}[/tex]
Yes
Explanation:
[tex]m_8[/tex] = Mass of uranium 238 ion = [tex]3.95\times 10^{-25}\ \text{kg}[/tex]
[tex]m_5[/tex] = Mass of uranium 235 ion = [tex]3.9\times 10^{-25}\ \text{kg}[/tex]
v = Velocity of ions = [tex]3\times 10^5\ \text{m/s}[/tex]
q = Charge of triply charged ions = [tex]3\times 1.6\times 10^{-19}\ \text{C}[/tex]
B = Magnetic field = 0.25 T
The force balance is
[tex]\dfrac{mv^2}{r}=qvB\\\Rightarrow r=\dfrac{mv}{qB}[/tex]
The difference between the radius of the ions are
[tex]\Delta r=(m_8-m_5)\dfrac{v}{qB}\\\Rightarrow \Delta r=\dfrac{(3.95\times 10^{-25}-3.9\times 10^{-25})\times 3\times 10^5}{3\times 1.6\times 10^{-19}\times 0.25}\\\Rightarrow \Delta r=0.0125\ \text{m}[/tex]
Separation is given by
[tex]\Delta d=2\Delta r=2\times 0.0125\\\Rightarrow \Delta d=0.025\ \text{m}[/tex]
The separation between their paths when they hit a target after traversing a semicircle is [tex]0.025\ \text{m}[/tex].
Yes, the distance between the paths is 2.5 cm is a practical separation between as it is easily measurable.
