Problem 1

1.) (10 points) A ⁵⁸Ni ion has a charge of +e and a mass of $9.62 \times 10^{-26} kg$ is accelerated through a potential difference of 3 kV and deflected in a magnetic field of 0.12 T. It follows the path shown below.  
a) How far down from where the ion entered will it strike the wall? In other words, find the distance d.
(Hint: Think about conservation of energy)
 
b) We know that mixed in with our ⁵⁸Ni ions, we have a few ⁶⁰Ni ions. They have the same charge, just a greater mass. (Assume the mass ratio is excatly 58/60) How far from where the ⁵⁸Ni ions strike the wall do the ⁶⁰Ni ions strike?
 

$a_{centripetal} = \frac{v^{2}}{r}$
$U = q \Delta V$
$q_{electron} ~=~ -e ~=~ -1.6 \times 10^{-19} C$

 

Answer:
a)

Starting with the equation that
$U = q \Delta V$
We have that the potential energy of the particles entering the B field is given by
$U = (1.6 \times 10^{-19})(3000) ~=~ 4.8 \times 10^{-16} J$
Solving this for the velocity that the ions have when they enter the region of B field, we find that
$U = \frac{1}{2} m v^{2}$
$v = \sqrt{\frac{2 U}{m}}$
v = 99900 m/s
Now, we know that the force acting of the particles by the magnetic field will cause them to move in a circular orbit, so we see that
$F_{m} = m \frac{v^{2}}{r}$
$qvB = \frac{mv^{2}}{r}$
Solving for the radius, we find that
$r = \frac{mv}{qB}$
Putting our value for the velocity in to this equation, we see that
r = .501 m
So, the distance d = 2r is given by
d = 1.002 m

b)
Now the mass is given by
$m_{60} = \frac{60}{58} m_{58}$
$m_{60} = 9.95 \times 10^{-26} kg$
So, the velocity is now
v₆₀ = 98200 m/s
and the radius is
r₆₀ = .509 m
And the distance d₆₀ is given by
d₆₀ = 2r₆₀
d₆₀ = 1.017 m
So, the difference in where the ⁵⁸Ni ions hit and where the ⁶⁰Ni ions hit is given by
$\Delta d = d_{60}- d_{58}$
$\Delta d = .015 m$