s/mod/ibis/view.php?id 3196120 Dell 6 cmx Newegg a Amazon sapling f Facebook C c

Question

s/mod/ibis/view.php?id 3196120 Dell 6 cmx Newegg a Amazon sapling f Facebook C chegg study Rc Glossary Nc sides a lDriv g17-REIFENBERGERI Activities and Due Dates HW#4:Magnetism 017 09:00 AM (A 80.1/100 2/21/2017 1256 PM Print L Calculator Periodic Table 3 of 13 Sapling Learning 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 radioatictive similar storing both in bone marrow. The radium then bombards nearby bone cells with alpha particles, causing them to 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 shown is in a constant magnetic field of 0.652 T pointing out of the plane of the schematic. Motion of the positively-charged i toward the right was initiated by a potential difference of 2979 V on the two plates shown. Using the data shown in the table below, calculate the path radius of the Ca' ion. (Soroll down for table and more questions) Using the same data table, match the particles to their path label. A Previous 2 Check Answer 0 Next Exit

Explanation / Answer

First equation:

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=2979 V
m = 40.1amu*(1.661 × 10^-27 kg/1 amu)=6.661 x 10^-26 kg
B=0.652 T
and
r=?

Solve for r:

r=[2*1.602 × 10^-19 C*2979 V/6.661 x 10^-26 kg]/(1.602 × 10^-19 C*0.652 T/6.661 x 10^-26 kg )^2
r=[9.545 x 10^-16 C*V/6.66 x 10^-26 kg]/(10.45 x 10^-20 C*T/6.66 x 10^-26 kg)^2
r=1.433 x 10^10/2.46 x 10^12
r= 0.005825 m = 5.825 mm

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