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

Question

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)?

Answer 1

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

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

Radius of a semicircle in magnetic field

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

Finally, the mass of an ion

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

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

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