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

ID: 1349811 • Letter: A

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

m1 = 27 amu

m2 = 6 amu

m3 = N amu?


when a moving charged particle enters uniform magnetic field, it exeperiences magnetic force.

F = q*v*B

m*a = q*v*B

m*v^2/r = q*v*B

r = m*v/(B*q)


separtion between m1 and m2, 5.7 cm = 2*(r1 - r2)

= 2*(m1*v/(B*q) - m2*v/(B*q))

= 2*(v/(B*q))*(m1-m2)

= 2*(v/(B*q))*(27 - 6)*amu

5.7 cm = (21 amu *v/(B*q))

1 cm = 3.684 amu*v/(B*q)


now separtion between r2 and r3, r3 - r2 = 2*(m3*v/(B*q) - m2*v/(B*q))

= 2*(v/(B*q))*(m3-m2)

3.7 cm = 2*(v/(B*q))*(N - 8)*amu

3.7*3.684 amu*v/(B*q) = 2*(v/(B*q))*(N - 6)*amu

13.63*amu*v/(B*q) = 2*(v/(B*q))*(N - 6)*amu


13.63 = 2*(N - 6)

13.63/2 = N - 6

N = 6 + 13.63/2

= 12.82

so, m3 = 12.82 amu <<<<---------Answer

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