Question

Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions (Au^2+), each with a mass of 3.27*10^(-25) kg. The ions are accelerated from rest through a potential difference of 1.10 kV. Then, a 0.490 T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

Answer 1

Charge \((q)\) of the doubly ionized \(\mathrm{g}\) old \(=2 e\)

$$ \begin{aligned} &=2\left(1.6 \times 10^{-19} \mathrm{C}\right) \\ &=3.2 \times 10^{-19} \mathrm{C} \end{aligned} $$

Accelerating potential is \(V=1.10 \times 10^{3} \mathrm{~V}\)

Now speed \((v)\) of these ions as passing through this accelerating

potential is

$$ \begin{aligned} q V &=\frac{1}{2} m v^{2} \\ v &=\sqrt{\frac{2 q V}{m}} \\ &=\sqrt{\frac{2\left(3.2 \times 10^{-19} \mathrm{C}\right)\left(1.10 \times 10^{3} \mathrm{~V}\right)}{3.27 \times 10^{-25} \mathrm{~kg}}} \\ &=4.64 \times 10^{4} \mathrm{~m} / \mathrm{s} \end{aligned} $$

WHen these ions are passing through magnetic field then they

follow the circular path of radius is given by

$$ \begin{aligned} r &=\frac{m v}{B q} \\ &=\frac{\left(3.27 \times 10^{-25} \mathrm{~kg}\right)\left(4.64 \times 10^{4} \mathrm{~m} / \mathrm{s}\right)}{(0.490 \mathrm{~T})\left(3.2 \times 10^{-19} \mathrm{C}\right)} \\ &=9.68 \times 10^{-2} \mathrm{~m} \\ &=9.68 \mathrm{~cm} \end{aligned} $$