The region shown is immersed in a constant magnetic field of 0.552 T pointing ou
ID: 1322418 • Letter: T
Question
The region shown is immersed in a constant magnetic field of 0.552 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 2773 V on the two plates shown. Using the data shown in the table below, calculate the path radius of the Ca ion.
The region shown is immersed in a constant magnetic field of 0.552 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 2773 V on the two plates shown. Using the data shown in the table below, calculate the path radius of the Ca ion. Using the same data table, match the particles to their path label. 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, banum (another alkaline earth metal) and radium isotopes entering the chamber. The region shown is immersed in a constant magnetic field of 0.552 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 2773 V on the two plates shown. Using the data shown in the table below, calculate the path radius of the ca^+. ion.Explanation / Answer
Use KE = PE
0.5 mv^2 = eV
so speed v = sqrt(2eV/m)
apply centripetal force = magnetic force
i.e mv^2/r = qvB,
where m = mass o the charged particle
v = velocity, r = radius ,
q = charge = 1.6*10^-19 C
B = magnetic field
so now , so as r = mv/qB is the formula needed.
for Ca+ ion , m = 0.666 e -25 kgs
q = 1.602 e-19C
so
v = sqrt(2* 1.6 e-19 * 2773)/(0.666 e-25)
v = 1.154 e 5 m/s
so raidus of Ca+ = (0.666e-25* 1.154 e5/(1.6e-19 * 0.552)
radius r = 0.087 m
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also r = (m/qB) sqrt(2eV/m)
radius is directly proportioaal to ratio of m/q
so large m/q corresponds to large radius
so
for Ca+ : m/q = 0.666e-25/1.6e-19 = 0.416 um
for Ca2+: m/q = 0.666 e-25/3.204 e-19 = 0.207 um
for Ba+ : m/q = 2.28 e-25/1.603 e-19 = 1.422 um
for Ba2+, m/q = 2.28 e-25/3.204 e-19 = 0.711 um
for Ra+ , m/q = 3.75 e-25 /1.603 e-19 = 2.33 um
for Ra2+, m/q = 3.75 e-25/(3.204e-19) =1.17 um
for Ra3+. m/q = 3.75 e-25/4.806 e-19 = 0.78um
so
correct matching will be
path A is Ca2+
path B is Ca+
path C is Ba2+
path D is Ra3+
path E is Ra2+
path F is Ba+
path G is Ra+