Question

A helium leak detector uses a mass spectrometer to detect tiny leaks in a vacuum chamber. The chamber is evacuated with a vacuum pump and then sprayed with helium gas on the outside. If there is any leak, the helium molecules pass through the leak and into the chamber, whose volume is sampled by the leak detector. In the spectrometer, helium ions are accelerated and released into a tube, where their motion is perpendicular to an applied magnetic field, \(\vec{B},\) and they follow a circular orbit of radius \(r\) and then hit a detector. Estimate the velocity required if the orbital radius of the ions is to be no more than \(5 \mathrm{~cm},\) the magnetic field is \(0.15 \mathrm{~T}\) and the mass of a helium- 4 atom is about \(6.6 \cdot 10^{-27} \mathrm{~kg}\). Assume that each ion is singly ionized (has one electron less than the neutral atom). By what factor does the required velocity change if helium- 3 atoms, which have about \(\frac{3}{4}\) as much mass as helium- 4 atoms, are used?

Short answer

Answer: The estimated velocity required for helium-4 ions is 5.45 x 10^5 m/s. If helium-3 atoms are used instead, the required velocity will change by a factor of 4/3, meaning the required velocity will be larger for helium-3 ions.

Step-by-step solution

Formula for the motion of charged particles in a magnetic field

To determine the velocity of the helium ions, we need to use the formula for the motion of charged particles in a magnetic field: \(r = \frac{mv}{qB}\) where: \(r\) is the radius of the circular orbit \(m\) is the mass of the ion \(v\) is the velocity of the ion \(q\) is the charge of the ion \(B\) is the magnetic field strength

Calculate the charge of the helium ion

Since the problem states each helium ion is singly ionized, it means that each ion has lost one electron. The charge of an electron is \(1.6 \cdot 10^{-19} \mathrm{C}\), so the charge of a helium ion would be: \(q = 1.6 \cdot 10^{-19} \mathrm{C}\)

Rearrange the formula and plug in known values to solve for velocity

Rearrange the formula to solve for velocity: \(v = \frac{qrB}{m}\) Now plug in the known values for radius (\(r = 5 \cdot 10^{-2} \mathrm{m}\)), magnetic field strength (\(B = 0.15 \mathrm{T}\)), charge (\(q = 1.6 \cdot 10^{-19} \mathrm{C}\)), and mass of helium-4 (\(m = 6.6 \cdot 10^{-27} \mathrm{kg}\)): \(v^4 = \frac{(1.6 \cdot 10^{-19} \mathrm{C})(0.15 \mathrm{T})(5 \cdot 10^{-2} \mathrm{m})}{6.6 \cdot 10^{-27} \mathrm{kg}}\) Calculate the velocity of helium-4 ions: \(v^4 \approx 5.45 \times 10^5 \frac{\mathrm{m}}{\mathrm{s}}\)

Calculate the required velocity factor for helium-3 atoms

The mass of helium-3 is about \(\frac{3}{4}\) of the mass of helium-4. Let the velocity of helium-3 be represented as \(v^3\): \(v^3 = \frac{qrB}{m^3} = \frac{qrB}{\frac{3}{4}m}\) Now, we can find the ratio of the velocities (velocity factor) of helium-3 to helium-4: \(\frac{v^3}{v^4} = \frac{\frac{qrB}{\frac{3}{4}m}}{\frac{qrB}{m}}\) Simplify the equation and find the velocity ratio: \(\frac{v^3}{v^4} = \frac{4}{3}\)

State the final answer

The estimated velocity required for helium-4 ions to have an orbital radius of no more than \(5 \mathrm{~cm}\) in a magnetic field of \(0.15 \mathrm{~T}\) is \(5.45 \times 10^5 \frac{\mathrm{m}}{\mathrm{s}}\). If helium-3 atoms are used instead, the required velocity will change by a factor of \(\frac{4}{3}\), which means the required velocity will be larger for helium-3 ions.

Key concepts

Charged Particles in Magnetic Field

Whenever a charged particle, such as an ion, enters a magnetic field at an angle perpendicular to the field lines, it experiences a force. This force, known as the Lorentz force, causes the particle to move in a circular path.

The radius of this circular path is determined by the mass of the particle, its velocity, the charge it carries, and the strength of the magnetic field. The equation representing this relationship is given by the formula:
\r\( r = \frac{mv}{qB} \).

Here,

• \r\( r \) is the circular orbit's radius;
• \r\( m \) is the particle's mass;
• \r\( v \) is its velocity;
• \r\( q \) is its charge;
• \r\( B \) is the magnetic field strength.

In practical terms, for a given magnetic field and charge, lighter ions will follow tighter orbits at lower speeds, whereas heavier ions require either a greater speed or a larger orbit for the same magnetic field. This principle is fundamental in the operation of mass spectrometers, which can separate ions based on their mass-to-charge ratio.

Helium Ion Velocity Calculation

The calculation of the velocity of helium ions in a mass spectrometer involves rearranging the fundamental formula of motion in a magnetic field to make velocity the subject.

We use the rearranged form of the equation:
\r\( v = \frac{qrB}{m} \).

To find the velocity for a helium-4 atom with a known mass, charge, and magnetic field, we'd plug the known values into this formula. A key point to remember is that ions must be accelerated to the correct velocity to maintain the desired orbital radius within the spectrometer's magnetic field.

When calculating, it's essential to ensure that the units are consistent. For instance, the charge is typically given in coulombs (C), the magnetic field in teslas (T), the mass in kilograms (kg), and the velocity in meters per second (m/s).

The velocity calculation helps to ensure that the ions hit the detector at the expected point, allowing accurate identification of leaks in applications like helium leak detectors.

Mass-to-Charge Ratio

The mass-to-charge ratio (\r\( \frac{m}{q} \)) is a crucial aspect in the analysis of charged particles moving through a magnetic field, as it directly influences the trajectory of the particles within a mass spectrometer.

Ions with different mass-to-charge ratios will have different velocities for a given magnetic field strength and radius of curvature, leading to their separation within the spectrometer. This property is employed to distinguish between ions of different atomic or molecular species, making mass spectrometry a powerful analytical tool for everything from elemental analysis to biomolecular studies.

The exercise also introduces the variation of required velocity when using helium-3 atoms instead of helium-4. Since helium-3 has about three-quarters the mass of helium-4, the velocity needs to be greater for helium-3 to maintain the same radius. This adjustment, by a factor of \r\( \frac{4}{3} \), underscores how changing the mass affects the velocity in a mass spectrometer setup.