Answer :
Answer:
0.092 m
Explanation:
A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.
Since the magnetic force acts as centripetal force, we can write:
[tex]qvB=m\frac{v^2}{r}[/tex]
where
q is the charge of the particle
v is its velocity
B is the strength of the magnetic field
m is the mass of the particle
r is the radius of the orbit
Solving the equation for r,
[tex]r=\frac{mv}{qB}[/tex]
For the ion of oxygen-16, we have:
[tex]m_A=2.66\cdot 10^{-26}kg[/tex]
[tex]q_A = 1.6\cdot 10^{-19}C[/tex] (it is singly charged)
[tex]v_A=2.90\cdot 10^6 m/s[/tex]
[tex]B_A=1.30 T[/tex]
So the radius of its orbit is
[tex]r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m[/tex]
For the ion of oxygen-18, we have:
[tex]m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg[/tex]
[tex]q_B = 1.6\cdot 10^{-19}C[/tex] (it is singly charged)
[tex]v_B=2.90\cdot 10^6 m/s[/tex]
[tex]B_B=1.30 T[/tex]
So the radius of its orbit is
[tex]r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m[/tex]
After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:
[tex]d=2(r_B-r_A)=2(0.417-0.371)=0.092 m[/tex]