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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

Respuesta :

Answer:

0.132 m

Explanation:

[tex]m[/tex] = mass of the ion = 5.29 x 10⁻²⁷ kg

[tex]q[/tex] = magnitude of charge on singly charged ion = 1.6 x 10⁻¹⁹ C

[tex]r[/tex] = radius of circular path followed by singly charged ion

[tex]v[/tex] = speed of the ion = 1.13 x 10⁶ m/s

[tex]B[/tex] = magnitude of the magnetic field = 0.283 T

Radius of the circular path is given as

[tex]r = \frac{mv}{qB}[/tex]

[tex]r = \frac{(5.29\times 10^{-27})(1.13\times 10^{6})}{(1.6\times 10^{-19})(0.283)}[/tex]

[tex]r[/tex] = 0.132 m

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