How to Compare Ion Path Radii After the Same Accelerating Voltage

Two ions are accelerated through the same voltage and then enter the same magnetic field. One ion is four times as massive as the other. Does it follow a path with four times the radius?

No. If both ions start from rest and enter perpendicular to the magnetic field, the heavier ion follows a path with twice the radius, provided their charge magnitudes are equal.

The general comparison is:

\[ \boxed{r=\frac{\sin\theta}{B} \sqrt{\frac{2m\Delta V}{|q|}}} \]

This means the path radius depends on three things: the ion’s mass-to-charge ratio, the magnetic field strength, and the angle at which it enters the field.

For ions entering at the same angle, the larger \(m/|q|\) produces the larger radius. If their entry angles differ, you must account for the perpendicular component of velocity before comparing them.

One important clarification: the radius here is the radius of the ion’s curved trajectory, not the physical or chemical ionic radius.

01The experiment: from accelerating voltage to magnetic deflection

Imagine an experiment designed to separate ions according to their mass-to-charge ratio.

The apparatus contains five main parts:

1. An ion source produces charged particles.
2. Two accelerating plates create a potential difference, \(\Delta V\).
3. A narrow slit produces a more controlled beam of ions.
4. The beam enters a region containing a uniform magnetic field, \(B\).
5. A detector records where the ions arrive.

The accelerating voltage gives the ions kinetic energy. The magnetic field then bends their trajectories, allowing different ions to reach different positions on the detector.

The experiment is trying to answer a specific question:

Can we use the radius of an ion’s curved path to distinguish ions with different masses or charges?

Before deriving the relationship, consider what you would expect.

Suppose two singly charged ions enter the same magnetic field at the same angle. Ion B has four times the mass of ion A.

You might predict that B has four times the radius. After all, it is four times as massive.

The actual result is that B has twice the radius.

The reason is that accelerating through the same voltage does not give particles of different masses the same speed. We need to understand both the energy gained and the magnetic force acting on the ions.

02Why only the perpendicular velocity component matters

An ion entering a magnetic field experiences a magnetic force:

\[ F_B=|q|vB\sin\theta \]

Here:

• \(F_B\) is the magnitude of the magnetic force, in newtons (N).
• \(q\) is the signed electric charge, in coulombs (C).
• \(v\) is the ion’s total speed, in metres per second (m/s).
• \(B\) is the magnetic field strength, in tesla (T).
• \(\theta\) is the angle between the velocity and magnetic field.

The key part is \(\sin\theta\).

This tells us that the magnetic force depends on the component of velocity perpendicular to the magnetic field.

We call this the perpendicular velocity component:

\[ v_\perp=v\sin\theta \]

There is also a parallel component:

\[ v_\parallel=v\cos\theta \]

The parallel component does not experience a magnetic force.

This gives us two different types of motion.

Perpendicular motion: The magnetic force acts as a centripetal force, continuously changing the direction of the perpendicular velocity.

Parallel motion: The ion continues moving along the magnetic field direction at constant velocity.

When the ion enters perpendicular to the field, there is no parallel component. Its trajectory is circular.

When it enters at another angle, the combination of circular motion and parallel motion produces a helix.

This distinction is important because the radius of the circular part of the trajectory depends on \(v_\perp\), not the total speed \(v\).

Deriving the radius of the curved path

For circular motion, the magnetic force supplies the centripetal force.

The magnetic force is:

\[ F_B=|q|v_\perp B \]

The centripetal force is:

\[ F_c=\frac{mv_\perp^2}{r} \]

Equating the two:

\[ |q|v_\perp B=\frac{mv_\perp^2}{r} \]

Rearranging:

\[ \boxed{r=\frac{mv_\perp}{|q|B}} \]

This is the starting equation for comparing ion trajectories.

Notice that the charge magnitude appears in the denominator. A larger charge magnitude produces a smaller radius if the other quantities remain constant.

Also, the equation uses the perpendicular velocity component. If an ion enters at an angle, using its total velocity without resolving the components gives the wrong radius.

03Why accelerating through the same voltage matters

Now we need to connect the magnetic motion to the accelerating voltage.

When a charged particle moves through a potential difference, electrical potential energy is converted into kinetic energy.

The kinetic energy gained is:

\[ \Delta E_k=|q|\Delta V \]

For an ion starting from rest:

\[ \frac12 mv^2=|q|\Delta V \]

Rearranging for speed:

\[ v=\sqrt{\frac{2|q|\Delta V}{m}} \]

This equation reveals something important.

For the same accelerating voltage and charge magnitude:

• A heavier ion gains the same kinetic energy.
• The heavier ion reaches a lower speed.
• Its momentum is nevertheless greater.

That last point explains why the heavier ion can follow a wider curve despite moving more slowly.

Substitute the speed equation into the magnetic radius equation:

\[ r=\frac{mv\sin\theta}{|q|B} \]

Therefore:

\[ \boxed{ r=\frac{\sin\theta}{B} \sqrt{\frac{2m\Delta V}{|q|}} } \]

This is the main equation for comparing ions accelerated through the same voltage.

The comparison rule

If two ions start from rest and experience the same accelerating voltage and magnetic field:

\[ \frac{r_1}{r_2} = \frac{\sin\theta_1}{\sin\theta_2} \sqrt{ \frac{m_1/|q_1|}{m_2/|q_2|} } \]

If both enter perpendicular to the field, their angles are \(90^\circ\), so:

\[ \sin90^\circ=1 \]

The equation simplifies to:

\[ \boxed{ \frac{r_1}{r_2} = \sqrt{ \frac{m_1/|q_1|}{m_2/|q_2|} } } \]

This is the relationship to remember when comparing ion path radii.

For ions with the same charge magnitude, it becomes:

\[ \frac{r_1}{r_2}=\sqrt{\frac{m_1}{m_2}} \]

A useful summary:

04Identifying the charge sign from curvature

The radius tells us about the mass-to-charge magnitude. It does not tell us whether the ion is positive or negative.

For that, we need the direction of the magnetic force.

The magnetic force on a charged particle is:

\[ \vec F_B=q(\vec v\times\vec B) \]

The direction depends on three things:

• The direction of the particle’s velocity.
• The direction of the magnetic field.
• The sign of its charge.

Consider a simple arrangement.

The magnetic field points into the page, represented by crosses (×). The ion enters moving horizontally to the right.

Using the right-hand rule for a positive charge:

1. Point your fingers in the direction of velocity.
2. Curl them towards the magnetic field direction.
3. Your thumb gives the direction of the magnetic force.

The positive ion experiences an upward force and curves upwards.

A negative ion experiences the opposite force and curves downwards.

There are two important rules here:

1. Opposite charge signs produce opposite curvature directions.

If the velocity and magnetic field directions are known, the curvature tells you the charge sign.

2. Opposite charge signs do not produce different radius magnitudes solely because of their signs.

The radius equation contains \(|q|\), not signed \(q\). Changing the charge from positive to negative reverses the direction of curvature but does not change the radius if all other conditions are identical.

One final trap: a straight trajectory does not automatically mean the particle is neutral. A charged particle travelling exactly parallel to the magnetic field experiences no magnetic force either.

05Worked examples

The following examples use illustrative values. Take:

\[ u=1.66\times10^{-27}\text{ kg} \]

\[ e=1.60\times10^{-19}\text{ C} \]

Here, \(u\) is the atomic mass unit and \(e\) is the elementary charge magnitude.

06Why the experimental setup needs to be controlled

There is a limitation in the simple experiment.

Suppose ions enter a magnetic field with different initial speeds. Even if they have identical masses and charges, they can follow different trajectories.

A larger perpendicular velocity produces a larger radius.

Likewise, ions entering at different angles can produce different radii even if their mass-to-charge ratios are identical.

The original magnetic-only arrangement therefore cannot isolate mass-to-charge ratio unless the initial conditions are known.

A revised experiment controls these variables:

• Accelerating plates provide a known potential difference.
• A narrow slit or collimator controls the direction of the incoming beam.
• A uniform magnetic field provides a known magnetic force environment.
• A detector records the resulting trajectories.

There is another subtlety. Accelerating ions from rest through the same voltage controls their kinetic energy gain per unit charge. It does not remove any initial kinetic energy they already possess.

If the ions have significant or different initial kinetic energies, the simple comparison equation needs to be modified.

A velocity selector can also be added to a mass spectrometer. It uses crossed electric and magnetic fields to select particles with a particular speed. In that arrangement, the comparison uses:

\[ r=\frac{mv}{|q|B} \]

because the speed is selected directly rather than determined by the accelerating voltage.

These are two different experimental approaches. Be careful not to mix their assumptions.

The voltage-based equation also assumes non-relativistic speeds, negligible collisions, and a sufficiently uniform magnetic field.

07Questions and solutions

All numerical values below are illustrative.

Question 1

Two positive, singly charged ions, A and B, start from rest and are accelerated through the same potential difference. They enter the same uniform magnetic field at \(90^\circ\) to the field.

Their masses are:

• Ion A: 25 u
• Ion B: 100 u

If ion A follows a circular path of radius 3.0 cm, calculate the radius of ion B’s path.

Solution 1

Ion B follows a path with a radius of 6.0 cm, twice the radius of ion A.

Both ions have the same charge magnitude, accelerating voltage, magnetic field, and entry angle.

Therefore:

\[ \frac{r_B}{r_A}=\sqrt{\frac{m_B}{m_A}} \]

Substitute:

\[ \frac{r_B}{3.0} = \sqrt{\frac{100}{25}} \]

\[ \frac{r_B}{3.0}=\sqrt4=2 \]

Therefore:

\[ \boxed{r_B=6.0\text{ cm}} \]

The mass increases by a factor of four, but the radius only increases by a factor of two.

Remember that the relationship involves the square root of the mass-to-charge ratio, not the ratio itself.

Question 2

Two ions, X and Y, start from rest and are accelerated through the same potential difference. They enter the same magnetic field perpendicular to the field, with both velocity vectors pointing horizontally to the right.

The magnetic field points into the page.

Their masses are:

• Ion X: 36 u
• Ion Y: 18 u

A detector shows that both ions follow paths with equal radii.

Ion X initially curves upwards, while ion Y initially curves downwards.

Determine:

a) The charge sign of each ion.

b) The ratio of their charge magnitudes, \(|q_X|:|q_Y|\).

Explain why the heavier ion does not have a larger radius.

Solution 2

Ion X is positive, ion Y is negative, and the charge magnitude of X is twice that of Y.

Part a: Determine the charge signs.

For a positive ion travelling to the right in a magnetic field directed into the page, the right-hand rule gives an upward magnetic force.

Therefore, X must be positively charged.

Y curves in the opposite direction, so it must be negatively charged.

Part b: Compare the charge magnitudes.

The radius equation is:

\[ r=\frac{1}{B} \sqrt{\frac{2m\Delta V}{|q|}} \]

Since the radii, magnetic fields, and accelerating voltages are equal:

\[ \frac{m_X}{|q_X|} = \frac{m_Y}{|q_Y|} \]

Rearranging:

\[ \frac{|q_X|}{|q_Y|} = \frac{m_X}{m_Y} \]

Substitute:

\[ \frac{|q_X|}{|q_Y|} = \frac{36}{18} =2 \]

Therefore:

\[ \boxed{|q_X|:|q_Y|=2:1} \]

The heavier ion has twice the charge magnitude, which compensates for its larger mass in the mass-to-charge ratio.

Equal radii do not necessarily mean equal masses or equal charges.

Question 3

An ion with mass 32 u and charge \(+2e\) starts from rest and is accelerated through a potential difference of 1.5 kV.

It enters a uniform magnetic field of strength 0.30 T at an angle of \(30^\circ\) to the magnetic field.

Calculate:

a) The radius of its helical trajectory.

b) The fraction of its total kinetic energy associated with perpendicular motion.

c) The direction of its initial magnetic deflection if the magnetic field points into the page and its perpendicular velocity component points to the right.

Solution 3

The helical trajectory has a radius of approximately 3.72 cm, and 25% of the ion’s kinetic energy is associated with perpendicular motion. Its initial magnetic deflection is upwards.

Part a: Calculate the radius.

Convert the mass and charge:

\[ m=32(1.66\times10^{-27}) \]

\[ m=5.312\times10^{-26}\text{ kg} \]

\[ |q|=2(1.60\times10^{-19}) \]

\[ |q|=3.20\times10^{-19}\text{ C} \]

Use:

\[ r=\frac{\sin\theta}{B} \sqrt{\frac{2m\Delta V}{|q|}} \]

Substitute:

\[ r= \frac{\sin30^\circ}{0.30} \sqrt{ \frac{2(5.312\times10^{-26})(1.5\times10^3)} {3.20\times10^{-19}} } \]

The square-root term is approximately:

\[ 0.0223\text{ m} \]

Therefore:

\[ r=\frac{0.5}{0.30}(0.0223) \]

\[ \boxed{r=0.0372\text{ m}=3.72\text{ cm}} \]

Part b: Calculate the perpendicular kinetic energy fraction.

Kinetic energy is proportional to the square of velocity:

\[ E_k=\frac12mv^2 \]

The fraction associated with perpendicular motion is:

\[ \frac{E_{k,\perp}}{E_k} = \frac{v_\perp^2}{v^2} \]

Since \(v_\perp=v\sin\theta\):

\[ \frac{E_{k,\perp}}{E_k} =\sin^2\theta \]

Substitute:

\[ \frac{E_{k,\perp}}{E_k} =\sin^230^\circ \]

\[ =0.5^2 \]

\[ \boxed{25\%} \]

The remaining 75% is associated with motion parallel to the magnetic field.

Part c: Determine the deflection direction.

The ion is positively charged.

With its perpendicular velocity pointing right and the magnetic field pointing into the page, the right-hand rule gives an upward magnetic force.

Therefore, the ion initially curves upwards.

Question 4

Two unknown ions, A and B, start from rest and are accelerated through the same potential difference before entering the same uniform magnetic field.

Their entry angles are:

• Ion A: \(90^\circ\)
• Ion B: \(30^\circ\)

Both ions follow helical trajectories with a measured radius of 6.0 cm.

a) Calculate the ratio of their mass-to-charge magnitudes:

\[ \frac{m_B/|q_B|}{m_A/|q_A|} \]

b) If ion A has a mass of 40 u and a charge of \(+2e\), calculate the mass-to-charge magnitude of ion B in u/e.

c) If ion B is singly charged, determine its mass. Explain why its mass cannot be determined uniquely if its charge state is unknown.

Solution 4

Ion B has four times the mass-to-charge magnitude of ion A.

Its mass-to-charge magnitude is \(80\text{ u}/e\). If it is singly charged, its mass is 80 u.

Part a: Compare their mass-to-charge magnitudes.

Start with:

\[ r=\frac{\sin\theta}{B} \sqrt{\frac{2m\Delta V}{|q|}} \]

Rearranging the comparison equation:

\[ \frac{m_B/|q_B|}{m_A/|q_A|} = \left( \frac{r_B}{r_A} \right)^2 \left( \frac{\sin\theta_A}{\sin\theta_B} \right)^2 \]

The radii are equal, so:

\[ \frac{m_B/|q_B|}{m_A/|q_A|} = \left( \frac{\sin90^\circ}{\sin30^\circ} \right)^2 \]

\[ =\left(\frac{1}{0.5}\right)^2 \]

\[ \boxed{\frac{m_B/|q_B|}{m_A/|q_A|}=4} \]

Ion B has four times the mass-to-charge magnitude.

Part b: Calculate ion B’s mass-to-charge magnitude.

For ion A:

\[ \frac{m_A}{|q_A|} = \frac{40u}{2e} \]

\[ =20\frac{u}{e} \]

Therefore:

\[ \frac{m_B}{|q_B|} = 4(20) \]

\[ \boxed{\frac{m_B}{|q_B|}=80\frac{u}{e}} \]

Part c: Determine the mass if ion B is singly charged.

If ion B has charge magnitude \(e\):

\[ m_B=80\frac{u}{e}\times e \]

\[ \boxed{m_B=80u} \]

However, without knowing the charge state, the mass is not uniquely determined.

For example:

• If B has charge magnitude \(e\), its mass is 80 u.
• If B has charge magnitude \(2e\), its mass is 160 u.

Both possibilities have the same mass-to-charge magnitude.

This is why magnetic deflection alone does not always identify an ion’s mass.

Question 5

Two unknown ion beams, P and Q, start from rest and are accelerated through the same potential difference. They then enter the same uniform magnetic field, whose direction is into the page.

Both beams have the same initial perpendicular velocity direction, pointing to the right, but their entry angles to the magnetic field are unknown.

Both form helical trajectories.

The measured properties are:

The pitch is the distance travelled parallel to the magnetic field during one complete revolution.

Determine:

a) The entry angle of each ion relative to the magnetic field.

b) The ratio of their mass-to-charge magnitudes.

c) The charge sign of each ion.

d) Whether the results prove that P and Q have the same mass.

Solution 5

The entry angles are approximately \(36.9^\circ\) for P and \(53.1^\circ\) for Q. Both ions have the same mass-to-charge magnitude, but opposite charge signs. The results do not prove that they have the same mass.

This question is more subtle than simply comparing the two radii.

A tempting approach is to compare 3.0 cm with 4.0 cm and conclude that Q has the larger mass-to-charge ratio.

That conclusion is incorrect because the ions enter at different angles. Their perpendicular velocity components are different.

The pitch provides the missing information.

Part a: Determine the entry angles.

During one complete revolution, the ion travels a distance \(p\) parallel to the magnetic field.

At the same time, its perpendicular motion completes a circle with circumference:

\[ 2\pi r \]

Both motions take the same time, so:

\[ \frac{p}{2\pi r} = \frac{v_\parallel}{v_\perp} \]

Using the velocity components:

\[ \frac{v_\parallel}{v_\perp} = \frac{v\cos\theta}{v\sin\theta} \]

Therefore:

\[ \boxed{\frac{p}{2\pi r}=\cot\theta} \]

For ion P:

\[ \cot\theta_P = \frac{8\pi}{2\pi(3)} \]

\[ \cot\theta_P=\frac43 \]

Therefore:

\[ \tan\theta_P=\frac34 \]

\[ \boxed{\theta_P=36.9^\circ} \]

For ion Q:

\[ \cot\theta_Q = \frac{6\pi}{2\pi(4)} \]

\[ \cot\theta_Q=\frac34 \]

Therefore:

\[ \tan\theta_Q=\frac43 \]

\[ \boxed{\theta_Q=53.1^\circ} \]

The two entry angles are complementary.

Part b: Compare the mass-to-charge magnitudes.

We know:

\[ r=\frac{mv_\perp}{|q|B} \]

Also, from the helical motion:

\[ \frac{p}{2\pi} = \frac{mv_\parallel}{|q|B} \]

Square both expressions and add them:

\[ r^2+\left(\frac{p}{2\pi}\right)^2 = \left(\frac{m}{|q|B}\right)^2 (v_\perp^2+v_\parallel^2) \]

Since:

\[ v_\perp^2+v_\parallel^2=v^2 \]

we obtain:

\[ r^2+\left(\frac{p}{2\pi}\right)^2 = \left(\frac{mv}{|q|B}\right)^2 \]

Using the energy equation for acceleration from rest:

\[ v^2=\frac{2|q|\Delta V}{m} \]

Substitution gives:

\[ \boxed{ \frac{m}{|q|} = \frac{B^2}{2\Delta V} \left[ r^2+\left(\frac{p}{2\pi}\right)^2 \right] } \]

This equation allows us to compare mass-to-charge magnitudes without knowing the entry angle.

For P:

\[ \ell_P= \sqrt{3^2+ \left(\frac{8\pi}{2\pi}\right)^2} \]

\[ \ell_P=\sqrt{9+16} \]

\[ \ell_P=5.0\text{ cm} \]

For Q:

\[ \ell_Q= \sqrt{4^2+ \left(\frac{6\pi}{2\pi}\right)^2} \]

\[ \ell_Q=\sqrt{16+9} \]

\[ \ell_Q=5.0\text{ cm} \]

Both have the same combined length.

Since the magnetic field and accelerating voltage are also identical:

\[ \boxed{ \frac{m_P}{|q_P|} = \frac{m_Q}{|q_Q|} } \]

Their mass-to-charge magnitudes are equal.

The larger radius of Q is balanced by its smaller parallel component of motion.

Part c: Identify the charge signs.

The magnetic field points into the page, and both ions initially have their perpendicular velocity components pointing to the right.

A positive ion experiences an upward magnetic force.

A negative ion experiences a downward magnetic force.

Since P curves upwards:

\[ \boxed{P\text{ is positive}} \]

Since Q curves downwards:

\[ \boxed{Q\text{ is negative}} \]

Part d: Do they have the same mass?

No. Equal mass-to-charge magnitudes do not establish equal masses.

For example, these two ions could have:

\[ \frac{20u}{e} \qquad\text{and}\qquad \frac{40u}{2e} \]

Both ratios are equal, even though their masses and charge magnitudes differ.

The experiment establishes that P and Q have equal mass-to-charge magnitudes, not that they are the same ion.

08The next step: mass-to-charge ratio or selected velocity?

When analysing charged particles, the most important decision is identifying what the experiment actually controls.

If ions start from rest and accelerate through the same voltage, use:

\[ r=\frac{\sin\theta}{B} \sqrt{\frac{2m\Delta V}{|q|}} \]

If a velocity selector gives the ions the same speed, use:

\[ r=\frac{mv_\perp}{|q|B} \]

Both equations describe magnetic deflection, but they rely on different experimental conditions.

For HSC Physics questions, start by identifying the accelerating voltage, magnetic field direction, entry angle, and charge sign. Once those are clear, you can choose the correct equation, compare the radii, and explain what the detector actually tells you.