Answer to Question #91947 in Electricity and Magnetism for Lawunn Khaing

Question #91947
The mass spectrometer is used to measure the mass of an ion. The initially stationary ion is accelerated by the electric field due to a potential difference P'and a uniform magnetic field is perpendicular to the path of the ion. A wide detector lines the bottom wall of the chamber, and the causes the lon to move in a semicircle and thus strike the detector. Suppose that B 80.0 mT, V 1000.0 V, and ions of charge 4 +1.6 x 10 C strike the detector at a point that lies at x = 1.6254 m. What is the mass m of the individual ions, in atomic mass units (1 u 1.6605 x 10 kg)?
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Expert's answer
2019-07-24T15:52:21-0400

The velocity of an ion after acceleration due to the electric field


v=2qVmv=\sqrt{\frac{2qV}{m}}

Radius of a semicircle in magnetic field

R=mvqB=2mVqB2R=\frac{mv}{qB}=\sqrt{\frac{2mV}{qB^2}}

Finally, the mass of an ion

m=qR2B22Vm=\frac{qR^2B^2}{2V}

=4×1.6×1019×(1.6254/2)2×(80×103)22×1000=\frac{4\times 1.6\times 10^{-19}\times(1.6254/2)^2\times (80\times 10^{-3})^2}{2\times 1000}

=13.52×1025kg=82u=13.52\times 10^{-25}\:\rm{kg}=82\:\rm{u}


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