In a mass spectrometer chlorine ions of mass 35u and charge +5e are emitted from a source and accelerated through a potential difference of

In a mass spectrometer chlorine ions of mass 35u and charge +5e are emitted from a source and accelerated through a potential difference of 250 kV. They then enter a region with a magnetic field which is perpendicular to their original direction of motion. The chlorine ions exit the spectrometer after being bent along a path with radius of curvature 3.5 m. What is the value of the magnetic field? (u = 1.66

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  1. Answer:

    Magnetic field will be equal to 0.269 T

    Explanation:

    We have given mass of chlorine ion is 35u

    As we know that 1 u = [tex]=1.66\times 10^{-27}kg[/tex]

    Radius of circular path is given r = 3.5 m

    So mass of chlorine ion [tex]=35\times 1.66\times 10^{-27}kg=58.1\times 10^{-27}kg[/tex]

    Charge [tex]q=1.6\times 10^{-19}C[/tex]

    Potential difference V = 250 KV

    From conservation of energy

    [tex]\frac{1}{2}mv^2=qV[/tex]

    So [tex]\frac{1}{2}\times 58.1\times 10^{-27}\times v^2=1.6\times 10^{-19}\times 250000[/tex]

    [tex]v=2.6\times 10^6m/sec[/tex]

    We know that radius is equal to [tex]r=\frac{mv}{qB}[/tex]

    So [tex]3.5=\frac{58.1\times 10^{-27}\times 2.6\times 10^6}{1.6\times 10^{-19}\times B}[/tex]

    [tex]B=0.269Tesla[/tex]

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