Question

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.60 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.00 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B},\) acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0}\), of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\). d) The gold ions (represented by the black circles) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=\) \(3.27 \cdot 10^{-25}\) kg. Calculate the mass of a molybdenum ion.

Short answer

Question: Determine the mass of a molybdenum ion in a mass spectrometer, given that the ions pass through a velocity selector and that the gold and molybdenum ions exit the particle detector at distances of 40.00 cm and 19.81 cm, respectively. Answer: The mass of a molybdenum ion is \(1.62 \cdot 10^{-25}\) kg.

Step-by-step solution

Determine the Directions of Electric and Magnetic Forces

To determine the directions of the electric and magnetic forces, first recall that the electric force is given by \(\vec{F}_{E} = q\vec{E}\), and magnetic force is given by \(\vec{F}_{B} = q\vec{v} \times \vec{B}\). Since the ion has a charge of \(+e\), inside the velocity selector, the electric force will be in the direction of the electric field, pointing upwards. The magnetic force, due to the cross product, will be perpendicular to both the velocity vector and the magnetic field vector. As the magnetic field is pointing out of the page, the magnetic force vector will be pointing either right or left, depending on the direction of the ion's velocity. #b) Calculating the Velocity of the Ions#

Calculate the Velocity of Ions Passing Through Velocity Selector

In the velocity selector, the electric and magnetic forces must balance each other for the ions to pass through in a straight line. That is \(F_E=F_B\). So, \(qE = qvB_1\). We can solve for the velocity, \(v_0\), that the ions must have: \(v_{0}=\frac{qE}{qB_{1}}=\frac{1.60 \cdot 10^{-19} \text{C} \times 1.789 \cdot 10^{4} \text{V/m}}{1.60 \cdot 10^{-19} \text{C} \times 1.00 \text{T}} = 1.789 \cdot 10^{4} \text{m/s}\) As the ratio of the electric and magnetic fields determines \(v_0\) and both ions have the same charge, \(v_0\) will be the same for gold and molybdenum ions. #c) Writing the Equation for the Radius of the Semicircular Path#

Write the Formula for the Radius of the Semicircular Path

The formula for the radius of the semicircular path of an ion in a magnetic field is given by: \(R = \frac{mv}{qB}\) where \(m\) is the ion's mass, \(v\) is its velocity, \(q\) is its charge, and \(B\) is the magnitude of the magnetic field. Thus, we can write the equation for the radius of the semicircular path in the particle detector as: \(R = R(m, v_0, q, B_2)\) #d) Calculating the Mass of a Molybdenum Ion#

Use the Radius Formula to Calculate the Mass of a Molybdenum Ion

We know that the gold ions exit the particle detector at a distance \(d_2=40.00 \text{cm}\) and the molybdenum ions exit at a distance \(d_1=19.81\text{cm}\). Since the path is a semicircle, the diameter is equal to the distance, so: \(R_\text{gold}=\frac{d_2}{2}=\frac{40.00\text{cm}}{2}=20.00\text{cm}=0.200\text{m}\) \(R_\text{molybdenum}=\frac{d_1}{2}=\frac{19.81\text{cm}}{2}=9.905\text{cm}=0.09905\text{m}\) Now we can use the radius formula to find the mass of a molybdenum ion: \(\frac{m_\text{molybdenum}}{m_\text{gold}}=\frac{R_\text{molybdenum}}{R_\text{gold}}\) Solve for \(m_\text{molybdenum}\): \(m_\text{molybdenum}=m_\text{gold}\times \frac{R_\text{molybdenum}}{R_\text{gold}}=3.27 \cdot 10^{-25}\text{kg}\times \frac{0.09905\text{m}}{0.200\text{m}}=1.62 \cdot 10^{-25}\text{kg}\) The mass of a molybdenum ion is \(1.62 \cdot 10^{-25}\) kg.

Key concepts

Understanding the Velocity Selector

The velocity selector is a clever device used in a mass spectrometer to ensure that only particles with a specific velocity passes through to the next stage of the spectrometry process. It uses both electric and magnetic fields positioned perpendicular to each other. Particles entering the velocity selector experience two forces: the electric force, which pushes them in the direction of the electric field, and the magnetic force, which is experienced when a charged particle moves through a magnetic field and acts perpendicular to both the magnetic field and the particle's velocity.
The key to the velocity selector's function is balance. When the electric force equals the magnetic force, the forces cancel each other out, meaning that particles continue their path undeviated. This occurs when the velocity of the particle is such that

\(F_E = F_B\) or \(qE = qvB\).
This magic velocity, which we denote as \(v_0\), depends solely on the magnitudes of the electric field (\(E\)) and the magnetic field (\(B\)), and it is uniform for all ions irrespective of the type, as long as they have the same charge.

The Role of Magnetic Force in Mass Spectrometry

Magnetic force plays a pivotal role in the functioning of a mass spectrometer. It is the force exerted on a charged particle when it moves through a magnetic field. The relationship between magnetic force (\(\vec{F}_{B}\)), charge (\(q\)), particle velocity (\(\vec{v}\)), and the magnetic field strength (\(\vec{B}\)) is captured by the formula

\(\vec{F}_{B} = q\vec{v} \times \vec{B}\).
Crucially, this force causes charged particles like ions to travel in a curved path. For an ion to pass through the velocity selector, its velocity must be set such that the magnetic force balances exactly with the electrical force. The magnetic force, essentially, becomes a filter for particles based on their velocity, and subsequently, within the mass spectrometer's detector, based on their mass-to-charge ratio.

Ion Path Radius Calculation in a Magnetic Field

Once the ions with the correct velocity (\(v_0\)) have been selected, they will enter the detector portion of the mass spectrometer and their path will curve due to the magnetic force. The radius of this path, known as the ion path radius, is of particular interest as it relates to the particles' mass-to-charge ratio. The stronger the magnetic field or the higher the velocity, the larger the radius will be.

To calculate this ion path radius (\(R\)), the formula \(R = \frac{mv}{qB}\) is used, where \(m\) stands for mass, \(v\) for velocity, \(q\) for charge, and \(B\) for the magnetic field strength. This equation arises from equating the magnetic force (\(qvB\)) to the centripetal force required to keep the ion moving in a circle (\(\frac{mv^2}{R}\)), leading to the equation above. Hence, an ion's mass can be deduced when the radius of its path is measured, a concept that is widely utilized in mass spectrometry to identify unknown substances.