(#22R3) Thank you for helping me, I need your help! An ion that is doubly ionize
ID: 1376421 • Letter: #
Question
(#22R3)
Thank you for helping me, I need your help!
An ion that is doubly ionized passes through the velocity selector and into the deflection chamber of a mass spectrometer, as shown below. In the velocity selector the electric field has a magnitude of 8004 V/m, and the magnitude of the magnetic field in both the velocity selector and the deflection chamber is 0.0917 T. If in the deflection chamber the ion is detected at a distance of 11.9 cm from its entry point, determine the following.
(a) mass-to-charge ratio of the ion
kg/C
(b) mass of the ion
kg
(c) identity of the ion, assuming it's an element (Use only the masses of elements in their most common form as listed on the periodic table of elements.)
Explanation / Answer
E = electric field = 8004 V/m
B = magnetic field = 0.0917 T.
r = radius = diamter / 2 = 11.9 cm /2 = 5.95 cm = 0.0595 m
a) in the velocity selector , magnetic fforce on the ion is equal to the electric force.
so magnetic force = electric force
q V B Sin 90 = qE
V = E/B
V = 8004 / 0.0917 = 87284.624 m/s
in the deflection chamber , the radius ''r'' of semircircle traced by the ion is given as ::
r = mV/qB
so m/q = Br/V
m/q = (0.0917)(0.0595) / (87284.624)
m/q = 6.25 x 10-8 kg/C
b) charge on ion = q = 2 x 1.6 x 10-19 C = 3.2 x 10-19 C
so m/q = 6.25 x 10-8
m = 6.25 x 10-8 x 3.2 x 10-19
m = 2 x 10-26 kg
c) from the periodic table , the element is Carbon since 2 x 10-26 kg in amu is 12.044.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.