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

Question

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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Amity 6 months 2021-08-12T14:13:50+00:00 1 Answers 15 views 0

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    2021-08-12T14:15:35+00:00

    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 = =1.66\times 10^{-27}kg

    Radius of circular path is given r = 3.5 m

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

    Charge q=1.6\times 10^{-19}C

    Potential difference V = 250 KV

    From conservation of energy

    \frac{1}{2}mv^2=qV

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

    v=2.6\times 10^6m/sec

    We know that radius is equal to r=\frac{mv}{qB}

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

    B=0.269Tesla

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