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Two ions with masses of 5.29×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is 0.283 T. Each has a speed of 1.13 × 106 m/s, but one ion is singly charged and the other is doubly charged. Find the radius of the circular path followed by the singly charged ion in the field

User Malks
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1 Answer

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Answer:

0.132 m

Step-by-step explanation:


m = mass of the ion = 5.29 x 10⁻²⁷ kg


q = magnitude of charge on singly charged ion = 1.6 x 10⁻¹⁹ C


r = radius of circular path followed by singly charged ion


v = speed of the ion = 1.13 x 10⁶ m/s


B = magnitude of the magnetic field = 0.283 T

Radius of the circular path is given as


r = (mv)/(qB)


r = ((5.29* 10^(-27))(1.13* 10^(6)))/((1.6* 10^(-19))(0.283))


r = 0.132 m

User Balaji Gunasekar
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