Answer:
[tex]0.0084575\ \text{T}[/tex]
[tex]0.272\ \text{m}[/tex]
2.2 cm easily observable
Explanation:
[tex]m_1[/tex] = Mass of 12 C = [tex]1.99\times 10^{-26}\ \text{kg}[/tex]
[tex]m_2[/tex] = Mass of 13 C = [tex]2.16\times 10^{-26}\ \text{kg}[/tex]
[tex]r_1[/tex] = Radius of 12 C = [tex]\dfrac{25}{2}=12.5\ \text{cm}[/tex]
B = Magnetic field
v = Velocity of atom = 8.5 km/s
[tex]r_2[/tex] = Radius of 13 C
The force balance of the system is
[tex]qvB=\dfrac{m_1v^2}{r}\\\Rightarrow B=\dfrac{m_1v}{rq}\\\Rightarrow B=\dfrac{1.99\times 10^{-26}\times 8500}{12.5\times 10^{-2}\times 1.6\times 10^{-19}}\\\Rightarrow B=0.0084575\ \text{T}[/tex]
The required magnetic field is [tex]0.0084575\ \text{T}[/tex]
Radius is given by
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r\propto m[/tex]
So
[tex]\dfrac{r_2}{r_1}=\dfrac{m_2}{m_1}\\\Rightarrow r_2=\dfrac{m_2}{m_1}r_1\\\Rightarrow r_2=\dfrac{2.16\times 10^{-26}}{1.99\times 10^{-26}}\times 12.5\times 10^{-2}\\\Rightarrow r_2=0.136\ \text{m}[/tex]
The required diameter is [tex]2\times 0.136=0.272\ \text{m}[/tex]
Separation is given by
[tex]2(r_2-r_1)=2(0.136-0.125)=0.022\ \text{m}[/tex]
The distance of separation is 2.2 cm which is easily observable.