In a mass spectrometer, a singly ionized 24Mg ion has a mass equal to 3.983 10-26 kg and is accelerated through a 3.00-kV potential differen

Question

In a mass spectrometer, a singly ionized 24Mg ion has a mass equal to 3.983 10-26 kg and is accelerated through a 3.00-kV potential difference. It then enters a region where it is deflected by a magnetic field of 526 G. Find the radius of curvature of the ion’s orbit. Note: There are 10,000 G in 1 T and 1,000 V in 1 kV.

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Thiên Thanh 4 years 2021-07-14T05:01:54+00:00 1 Answers 19 views 0

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  1. minhkhue
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    2021-07-14T05:03:09+00:00
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    Answer:

    The radius of curvature of the ion’s orbit is 0.59 meters

    Explanation:

    Given that,

    Mass of the 24 Mg ion, m=3.983\times 10^{-26}\ kg

    Potential difference, V = 3 kV

    Magnetic field, B = 526 G

    Charge on single ionized ion, q=1.6\times 10^{-19}\ C

    The radius of the the path traveled  by the charge is circular. Its radius is given by :

    r=\dfrac{mv}{Bq}

    v is speed of particle.

    v can be calculated using conservation of energy as :

    \dfrac{1}{2}mv^2=qV\\\\v=\sqrt{\dfrac{2qV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^3}{3.983 \times 10^{-26}}} \\\\v=1.26\times 10^5\ m/s

    Radius,

    r=\dfrac{3.983 \times 10^{-26}\times 1.26\times 10^5}{0.0526\times 1.6\times 10^{-19}}\\\\r=0.59\ m

    So, the radius of curvature of the ion’s orbit is 0.59 meters.

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