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27 April, 15:00

Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of magnitude 1.29 T directed perpendicular to their velocities. (a) Determine the radius of their circular path. (Give your answer to three significant figures.)

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  1. 27 April, 15:01
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    Answer: 0.091 m

    Explanation:

    r = 1/B * √ (2mV/e), where

    r = radius of their circular path

    B = magnitude of magnetic field = 1.29 T

    m = mass of Uranium - 238 ion = 238 * amu = 238 * 1.6*10^-27 kg

    V = potential difference = 2.9 kV

    e = charge of the Uranium - 238 ion = 1.6*10^-19 C

    r = 1/1.29 * √[ (2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

    r = 1/1.29 * √ (2.21*10^-21 / 1.6*10^-19)

    r = 1/1.29 * √0.0138

    r = 1/1.29 * 0.117

    r = 0.091 m

    Therefore, the radius of their circular path is 0.091 m
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