A sample containing elements with atomic masses m 1 =27 and m 2 =6 and an unknow

Question

A sample containing elements with atomic masses m1=27 and m2=6 and an unknown element is placed into a mass spectrometer. The ions all have the same charge and are accelerated to the same initial velocity before they enter the magnetic field. On the surface of the detector of the spectrometer the lines corresponding to elements with m1and m2  are separated by separation1=5.7 cm. The unknown element is detected to the right of element with m2, separation2=3.7 cm from it.

What is the mass of the unknown element? The answer should be in "amu" units.

Explanation / Answer

Let the mass of unknown element = m3 amu

radius r2, of m2 is lowest, so its mass is minimum, m2=6 amu
radius r1, of m1 = 27 is right of m2, by a distance of 5.7 cm.
r1-r2=5.7cm
radius r3, of m3 is right of m2, by a distance of 3.7 cm
r3-r2 = 3.7cm

In a mass spectrometer,
mass/charge = constant*r^2

distance between mass m1 and m2 = separation1 = r1-r2 = 5.7 cm
distance between mass m2 and m3 = separation3 = r3-r2 = 3.7 cm

charges are same for masses m1,m2 and m3.

So mass = constant*r^2

r^2 = mass/constant = constant3*mass
r1-r2 = constant3*(m1^0.5-m2^0.5)
r3-r2 = constant3*(m3^0.5-m2^0.5)

(r1-r2)/(r3-r2) = (m1^0.5-m2^0.5)/(m3^0.5-m2^0.5)
(r1-r2)/(r3-r2) = 5.7/3.7 = 1.54
m1 = 27; m2 = 6;
1.54 = (m1^0.5-m2^0.5)/(m3^0.5-m2^0.5)
1.54 = (27^0.5-6^0.5)/(m3^0.5-6^0.5) = 2.75/(m3^0.5-2.45)
(m3^0.5-2.45) = 2.75/1.54 = 1.785
m3^0.5 = 1.785+2.45 = 4.235
m3 = 17.94 or 18 amu (approximately)

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