Constants PartA Singly ionized (one electron removed) atoms are accelerated and

Question

Constants PartA Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 152 V/m and the magnetic field is 3.17x10-2 T.The ions next enter a uniform magnetic field of magnitude 1.74x10-2 T that is oriented perpendicular to their velocity. How fast are the ions moving when they emerge from the velocity selector? u= 4790 m/s evious Answe Correct Part B If the radius of the path of the ions in the second magnetic field is 18.0 cm, what is their mass? m1.90.1025 kg Submit X Incorrect; Try Again; 4 attempts remaining

Explanation / Answer

In the velocity selector only ions with a particular velocity pass undeflected through the combined electric and magnetic fields of the selector..

the magnetic and electric field forces on the ions are equal and opposite

Bqv = Eq

selected velocity v = E/B

v = 152V/m / 3.17*10-2T

v = 4794.9526m/s

The magnetic force (Bqv) provides the centripetal force (mv²/r) for the ions ..
Bqv = mv²/r

m = Bqr / v

(3.17*10-2T)(1.60*10-19C)(0.18m) / (4794.9526m/s)

m = 1.904*10-25 kg

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