Question

Elements that appear in the same column of the periodic table often share similar chemical properties. In the case of the alkaline earth metals, this is troublesome since the body treats calcium (necessary for proper bone growth) and radium (a radioatictive element) as chemically similar, storing both in bone marrow. The radium then bombards nearby bone cells with alpha particles, causing them to \"crumble.\" Radium poisoning investigations often center on the identification of radium and its isotopes in bone samples using a mass spectrometer. Pictured is a schematic of a simplified mass spectrometer, showing the paths of calcium, barium (another alkaline earth metal) and radium isotopes entering the chamber. The region shown is immersed in a constant magnetic field of 0.352 T pointing out of the plane of the schematic. Motion of the positively-charged isotopes toward the right was initiated by a potential difference of 3082 V on the two plates shown. Using the data shown in the table below, calculate the path radius of the Ca ion.

Answer 1

1.) F=ma

2.) a=m*(v^2/r)

Substitute 2 into 1.

F=m*(v^2/r)

What is the force of a charge particle?


F=qVB => call this equation 3 and substitute into equation 1.

qvB=m*(v^2/r)

v's on both sides, and cancels out.

qB=m*(v/r)


We need to find v, so how do we find v?


A voltage was applied which accelerated the Ca ion. The voltage*charge of the ion gives us the potential energy lost and the amount of KE of the Ca ion. So, PE=q*V and KE=1/2mv^2, equate to each other and solve for v:


qV=1/2mv^2

sqrt*[2*qV/m]=v <= call this equation 4 and substitute it into are already substituted equation 1.  


qB=m*(sqrt*[2*qV/m]/r)

Now, we have to rearrange this equation and solve for r:


(qB/m)^2=[2*qV/m]/r)


r=[2*qV/m]/(qB/m)^2

Where

q=1.602 × 10^-19 C.
V=3082 V
m=40.1amu*(1.661 × 10^-27 kg/1 amu)=6.661 x 10^-26 kg  
B=0.352 T
and  
r=?

Solve for r:

r=[2*1.602 × 10^-19 C*3082 V/6.661 x 10^-26 kg]/(1.602 × 10^-19 C*0.352 T/6.661 x 10^-26 kg )^2

r=0.0205m