How Millikan’s Data Reveal Quantised Electric Charge
Learn how Millikan's oil-drop measurements revealed that electric charge exists in discrete units. Explore experimental data, identify integer multiples of the elementary charge, and work through HSC Physics calculations involving measurement uncertainty and systematic errors.
Imagine measuring the charge on ten different oil droplets. You don’t get ten identical results. Some droplets have more charge than others, and the measured values contain small differences caused by experimental uncertainty.
But look more closely. What if the charges repeatedly occur near \(1.60\), \(3.20\), \(4.80\), and \(6.40\times10^{-19}\ \text{C}\)?
That pattern is the important evidence. Millikan’s data revealed that electric charge is quantised because the measured charges on different oil droplets were consistent with whole-number multiples of one fundamental charge. Changes in the charge on individual droplets also occurred in discrete steps.
The conclusion came from repeated measurements, not from finding one unusually neat number.
01What was Millikan trying to establish?
At the beginning of the twentieth century, physicists were investigating the structure of atoms and the properties of electrons.
J. J. Thomson had already measured the electron’s charge-to-mass ratio, \(e/m_e\). However, this did not reveal the electron’s charge or mass separately.
Millikan’s oil-drop experiment provided an independent way to measure the elementary charge.
There was another important question to investigate:
Was electric charge continuous, meaning it could have any value, or did it exist in discrete units?
These two models make different predictions.
- Continuous-charge model: A charged droplet could have any charge magnitude within a range.
- Quantised-charge model: A charged droplet could only have a whole-number multiple of a fundamental charge.
The second model predicts:
\[
|q|=ne
\]
where:
- \(q\) is the net charge on the droplet, in coulombs (C).
- \(n\) is a positive integer.
- \(e\) is the magnitude of the elementary charge, approximately \(1.60\times10^{-19}\ \text{C}\).
For a negatively charged droplet, its signed charge is \(q=-ne\). For a positively charged droplet, \(q=+ne\).
The electron carries charge \(-e\), while the proton carries charge \(+e\).
02How did the experiment produce useful data?
Millikan needed to measure the charge on individual droplets rather than a large collection of droplets together.
His apparatus contained two horizontal metal plates, one above the other. An atomiser above the upper plate sprayed tiny oil droplets into the space between them. A small opening in the upper plate allowed droplets to enter the gap.
A lamp illuminated the droplets from the side, while a microscope or telescope allowed an individual droplet to be observed.
A variable voltage supply controlled the electric field between the plates.
The experimenter could follow an individual droplet as its motion changed.
For example:
- With the electric field switched off, the droplet fell under gravity.
- Its speed increased until air resistance balanced its effective weight.
- The electric field was switched on.
- The electric force opposed gravity, changing the droplet’s motion.
- The experimenter measured its motion and used the known experimental conditions to calculate its charge.
The droplets could also gain or lose charge when they interacted with ions or electrons in the surrounding air.
This allowed Millikan to observe changes in the charge of the same droplet over time.
The important point is that the microscope did not directly measure electric charge. It measured droplet position over time. The charge was calculated from the observed motion and the forces acting on the droplet.
For a fuller explanation of the apparatus, force balance, and the original measurement method, see Millikan’s Oil-Drop Experiment: HSC Physics Explained.
03From experimental observations to quantised charge
There are three important connections between what Millikan observed and what physicists could conclude.
Observation 1: Different droplets had different charges
Millikan measured the motion of many individual droplets.
The droplets did not all have the same charge. Some carried a greater net charge than others.
That was expected under either model. The important question was whether the different measurements followed a common pattern.
Observation 2: The measured charges followed a repeated numerical pattern
Consider the following illustrative measurements.
These values are invented for teaching purposes and are not Millikan’s original experimental data.
| Droplet | Measured charge magnitude (C) |
|---|---|
| A | \(1.60\times10^{-19}\) |
| B | \(3.20\times10^{-19}\) |
| C | \(4.80\times10^{-19}\) |
| D | \(6.40\times10^{-19}\) |
| E | \(8.00\times10^{-19}\) |
Look at the relationship between the measurements.
| Droplet | Charge in units of \(e\) |
|---|---|
| A | \(1e\) |
| B | \(2e\) |
| C | \(3e\) |
| D | \(4e\) |
| E | \(5e\) |
Every measurement is a whole-number multiple of the same fundamental charge.
This is exactly what the quantised-charge model predicts.
The evidence becomes stronger when many droplets are measured.
A single measurement could simply represent one particular charge value. Repeated measurements showing the same underlying increment provide evidence for a fundamental unit.
Observation 3: Charge changes occurred in discrete steps
There was another particularly useful observation.
Millikan could sometimes follow the same droplet while its charge changed.
For example, a negatively charged droplet might capture an additional electron. Its charge magnitude would increase by \(e\).
A sequence of charge states could therefore be:
\[
4e,\quad 5e,\quad 6e,\quad 7e
\]
The droplet was not gradually accumulating an arbitrary amount of charge. Its measured charge changed in steps consistent with the transfer of individual elementary charges.
This was important because it provided evidence about changes in charge, rather than relying only on comparisons between different droplets.
There is one condition to remember: when comparing successive measurements, the experiment must account for changes in droplet radius, the electric field, and other experimental conditions.
04Worked example: finding the elementary charge from experimental data
Consider an illustrative experiment in which one negatively charged oil droplet is followed through four successive charge states.
The droplet’s measured charge magnitudes are:
| State | Measured charge magnitude (C) |
|---|---|
| A | \(6.42\times10^{-19}\) |
| B | \(8.00\times10^{-19}\) |
| C | \(9.61\times10^{-19}\) |
| D | \(11.19\times10^{-19}\) |
The droplet undergoes three successive captures of one electron between these states.
Estimate the elementary charge and identify the number of excess electrons in each state.
Step 1
For a negatively charged droplet:
\[
|q|=ne
\]
The successive electron captures mean that the number of excess electrons increases by one between each state.
We therefore look for four consecutive integer values of \(n\).
Step 2
The measurements are consistent with:
| State | Likely \(n\) |
|---|---|
| A | 4 |
| B | 5 |
| C | 6 |
| D | 7 |
Step 3
Rearrange:
\[
e=\frac{|q|}{n}
\]
For state A:
\[
e_A=\frac{6.42\times10^{-19}}{4}
\]
\[
e_A=1.605\times10^{-19}\ \text{C}
\]
For state B:
\[
e_B=\frac{8.00\times10^{-19}}{5}
\]
\[
e_B=1.600\times10^{-19}\ \text{C}
\]
For state C:
\[
e_C=\frac{9.61\times10^{-19}}{6}
\]
\[
e_C=1.602\times10^{-19}\ \text{C}
\]
For state D:
\[
e_D=\frac{11.19\times10^{-19}}{7}
\]
\[
e_D=1.599\times10^{-19}\ \text{C}
\]
Step 4
All four estimates are close to:
\[
\boxed{e\approx1.60\times10^{-19}\ \text{C}}
\]
The small differences are consistent with experimental uncertainty.
The droplet therefore carries approximately four, five, six, and seven excess electrons in the respective states.
More importantly, the repeated charge increment provides evidence for quantisation.
05Why experimental uncertainty does not disprove quantisation
One of the most common mistakes when analysing Millikan’s data is expecting every calculated charge to be an exact integer multiple of \(e\).
Real measurements are not perfect.
The experiment relied on several measurements and assumptions, including:
- The radius of each oil droplet.
- The viscosity of the surrounding air.
- The electric field strength.
- The droplet’s terminal velocity.
- The effect of air resistance and buoyancy.
Small errors in these quantities affect the calculated charge.
For example, a calculated charge might be:
\[
q=4.83\times10^{-19}\ \text{C}
\]
This is not exactly equal to \(3e\), which is approximately:
\[
3e=4.80\times10^{-19}\ \text{C}
\]
However, the difference is only:
\[
0.03\times10^{-19}\ \text{C}
\]
If the experimental uncertainty is large enough to account for this difference, the measurement is consistent with three elementary charges.
The correct approach is to compare the measurements with the predicted integer multiples while considering their uncertainties.
Do not interpret a calculated value such as \(3.02e\) as evidence that a droplet contains 3.02 excess electrons.
For an ordinary oil droplet in this experiment, the net charge is an integer multiple of the elementary charge. A non-integer calculated result indicates limitations in the measurement or model.
Random errors and systematic errors
These two types of error affect experimental data differently.
| Error type | Effect on Millikan’s data |
|---|---|
| Random error | Produces scatter around the expected integer multiples. |
| Systematic error | Can shift all calculated charge values in the same direction. |
This distinction matters.
Random uncertainty might make a measurement slightly higher or lower than its expected value.
A systematic error, however, could make every calculated charge approximately 1% too small. The pattern of integer multiples might still be visible even though the estimated value of \(e\) is inaccurate.
Worked example: How can a systematic error affect the result?
Suppose an experimental calibration causes every calculated charge to be 0.6% lower than its true value.
The modern value of the elementary charge is:
\[
e=1.602\times10^{-19}\ \text{C}
\]
Calculate the reported value and explain whether the evidence for quantisation disappears.
Step 1
A 0.6% underestimate means the calculated value is 99.4% of the true value.
Therefore:
\[
e_{\text{measured}}=0.994e
\]
Step 2
\[
e_{\text{measured}}
=0.994(1.602\times10^{-19})
\]
\[
e_{\text{measured}}
=1.592\times10^{-19}\ \text{C}
\]
Step 3
The calculated elementary charge is slightly too small.
However, every droplet’s calculated charge has been reduced by the same percentage. The relationship between the different charge states remains consistent with integer multiples.
Therefore, the evidence for quantisation remains.
The experiment has a systematic calibration error affecting the accuracy of the numerical value, rather than necessarily invalidating the underlying physical conclusion.
06What happened to Millikan’s actual results?
Millikan’s historical results are a useful example of why experimental evidence and numerical accuracy must be considered separately.
In 1913, Millikan published an improved measurement of the elementary charge, reporting approximately:
\[
e=1.592\times10^{-19}\ \text{C}
\]
The modern SI value is:
\[
e=1.602176634\times10^{-19}\ \text{C}
\]
The modern value is now exact under the SI definition of the ampere. You can check it against the National Institute of Standards and Technology’s SI definitions.
Why was Millikan’s value slightly different?
Historical analysis identified an incorrect value for the viscosity of air as an important source of the discrepancy. Since viscosity affected the calculation of droplet size and charge, using an incorrect value shifted the final numerical result.
When the viscosity value was corrected, the calculated elementary charge agreed with an independent method involving the Faraday constant and Avogadro constant.
Sources: American Physical Society’s historical account of Millikan’s 1913 results and NIST’s historical review of fundamental constants.
There is also a historical discussion about which droplets Millikan included in his published results. Later examination of his laboratory notebooks revealed additional measurements, and historians have disagreed about how to interpret the exclusions.
This is a useful lesson in experimental science: researchers should explain which measurements were rejected, why they were rejected, and how those decisions affect their conclusions.
Millikan’s work was important not because every original measurement was perfect, but because the repeated pattern of discrete charge values provided strong evidence for a fundamental unit.
07How to distinguish strong evidence from a misleading result
When interpreting a question about Millikan’s data, separate the experimental observation from the physical conclusion.
| Experimental observation | Reasonable inference | Important limitation |
|---|---|---|
| Different droplets have different charges. | Droplets can carry different amounts of net charge. | Different charges alone do not establish quantisation. |
| Calculated charges cluster near integer multiples of one value. | Charge is consistent with a common elementary unit. | Measurement uncertainty must be considered. |
| One droplet changes charge in repeated steps. | Charge changes occur in discrete increments. | Changes in radius and experimental conditions must be controlled. |
| Different experiments give slightly different values of \(e\). | Calibration or modelling errors may affect accuracy. | A numerical discrepancy does not automatically disprove quantisation. |
A strong HSC response should make the connection between the observation and the inference explicit.
For example:
Millikan measured the charges on many individual oil droplets and found that the calculated values were consistent with whole-number multiples of approximately \(1.60\times10^{-19}\ \text{C}\). The repeated pattern, together with observations of discrete changes in the charge of individual droplets, provided evidence that electric charge is quantised.
Notice that this explains what was measured, what pattern appeared, and what conclusion could be drawn.
One final distinction is worth remembering. Millikan investigated quantisation of electric charge. Flame tests provide evidence for quantisation of electron energy levels. Both involve discrete values, but they describe different physical quantities. See Flame Tests and Quantised Energy Levels for HSC Chemistry for the chemistry connection.
08Questions and solutions
Question 1
An oil droplet has an experimentally determined charge magnitude of:
\[
q=8.01\times10^{-19}\ \text{C}
\]
Using:
\[
e=1.60\times10^{-19}\ \text{C}
\]
Calculate the likely number of excess electrons on the negatively charged droplet.
Explain why the calculated value does not need to be an exact integer before considering experimental uncertainty.
Solution 1
The droplet most likely has five excess electrons.
The relationship between charge and the elementary charge is:
\[
|q|=ne
\]
Rearranging:
\[
n=\frac{|q|}{e}
\]
Substituting:
\[
n=\frac{8.01\times10^{-19}}
{1.60\times10^{-19}}
\]
\[
n=5.006
\]
The calculated value is approximately 5.
Therefore, the droplet has five excess electrons.
The slight difference from exactly 5 is consistent with experimental uncertainty. It does not mean that the droplet contains a fractional number of excess electrons.
Question 2
An experiment produces the following charge measurements for four different oil droplets.
| Droplet | Charge magnitude (\(\times10^{-19}\ \text{C}\)) | Measurement uncertainty |
|---|---|---|
| A | 3.23 | ±0.04 |
| B | 4.78 | ±0.05 |
| C | 6.42 | ±0.04 |
| D | 9.57 | ±0.05 |
a. Determine the likely number of elementary charges on each droplet.
b. Calculate an estimate of the elementary charge using all four measurements.
c. Explain why the absence of a measurement near \(5e\) does not disprove charge quantisation.
Solution 2
The measurements are consistent with two, three, four, and six elementary charges, giving an estimated elementary charge of approximately \(1.60\times10^{-19}\ \text{C}\).
Part a: Identify the integer multiples
Comparing the measurements with likely multiples of \(e\):
| Droplet | Likely number of charges, \(n\) |
|---|---|
| A | 2 |
| B | 3 |
| C | 4 |
| D | 6 |
Part b: Estimate the elementary charge
Use:
\[
e=\frac{|q|}{n}
\]
For droplet A:
\[
e_A=\frac{3.23\times10^{-19}}{2}
\]
\[
e_A=1.615\times10^{-19}\ \text{C}
\]
For droplet B:
\[
e_B=\frac{4.78\times10^{-19}}{3}
\]
\[
e_B=1.593\times10^{-19}\ \text{C}
\]
For droplet C:
\[
e_C=\frac{6.42\times10^{-19}}{4}
\]
\[
e_C=1.605\times10^{-19}\ \text{C}
\]
For droplet D:
\[
e_D=\frac{9.57\times10^{-19}}{6}
\]
\[
e_D=1.595\times10^{-19}\ \text{C}
\]
Taking the mean:
\[
e_{\text{mean}}
=\frac{1.615+1.593+1.605+1.595}{4}
\times10^{-19}
\]
\[
\boxed{e_{\text{mean}}=1.602\times10^{-19}\ \text{C}}
\]
The result is consistent with the accepted value.
Part c: Explain the missing measurement
There is no requirement for every possible charge state to appear in the collected data.
The droplets may acquire different numbers of electrons, and the experiment only measures the particular droplets that are successfully observed.
The important evidence is that the measured values are consistent with whole-number multiples of one common elementary charge.
The absence of a droplet carrying \(5e\) does not contradict that relationship.
Question 3
A negatively charged oil droplet is observed through four successive charge states.
The measured charge magnitudes are:
\[
4.90,\quad6.50,\quad8.10,\quad9.70
\]
All values are in units of \(10^{-19}\ \text{C}\).
The droplet captures one additional electron between each successive state.
However, the charge-measurement apparatus has a constant positive zero offset of:
\[
0.10\times10^{-19}\ \text{C}
\]
A student calculates the elementary charge by dividing each uncorrected measurement by the corresponding number of excess electrons. The student concludes that the elementary charge is not constant.
Evaluate this conclusion.
Solution 3
The student’s conclusion is incorrect. The measurements are consistent with a constant elementary charge of \(1.60\times10^{-19}\ \text{C}\) once the systematic zero offset is removed.
The droplet undergoes three successive electron captures.
Therefore, the likely charge states are:
\[
3e,\quad4e,\quad5e,\quad6e
\]
Step 1: Correct the measurements
The apparatus adds a constant positive offset, so subtract it from every measurement.
The corrected measurements are:
\[
4.90-0.10=4.80
\]
\[
6.50-0.10=6.40
\]
\[
8.10-0.10=8.00
\]
\[
9.70-0.10=9.60
\]
All values remain in units of \(10^{-19}\ \text{C}\).
Step 2: Calculate the elementary charge
For the first state:
\[
e=\frac{4.80}{3}\times10^{-19}
\]
\[
e=1.60\times10^{-19}\ \text{C}
\]
For the second state:
\[
e=\frac{6.40}{4}\times10^{-19}
\]
\[
e=1.60\times10^{-19}\ \text{C}
\]
The remaining two states give the same result:
\[
\frac{8.00}{5}=1.60
\]
\[
\frac{9.60}{6}=1.60
\]
Therefore:
\[
\boxed{e=1.60\times10^{-19}\ \text{C}}
\]
Step 3: Explain the original mistake
The constant zero offset affects every measurement by the same absolute amount.
Dividing by different integer values does not remove that offset. In fact, the error contributes a different amount to each calculated estimate of \(e\).
Interestingly, subtracting consecutive uncorrected measurements also gives:
\[
6.50-4.90=1.60
\]
\[
8.10-6.50=1.60
\]
\[
9.70-8.10=1.60
\]
The constant offset cancels when successive measurements are subtracted.
This shows that discrete spacing can remain visible even when the absolute charge calibration is incorrect.
Question 4
An experiment produces an inferred droplet charge of:
\[
q=5.44\times10^{-19}\ \text{C}
\]
with a stated uncertainty of:
\[
\pm0.04\times10^{-19}\ \text{C}
\]
Other droplets in the experiment produce measurements consistent with:
\[
e=1.60\times10^{-19}\ \text{C}
\]
The unusual droplet is exceptionally small, and its observed motion is irregular due to noticeable Brownian motion.
A student concludes:
The droplet has 3.4 excess electrons, so charge cannot be quantised.
Evaluate the student’s reasoning. Identify the likely experimental problems and suggest appropriate improvements.
Solution 4
The student cannot conclude that charge is not quantised. The measurement is inconsistent with the expected integer multiples, but the droplet’s irregular motion indicates that the simplified measurement model may not be reliable.
First, calculate the expected charge values.
For three elementary charges:
\[
q_3=3(1.60\times10^{-19})
\]
\[
q_3=4.80\times10^{-19}\ \text{C}
\]
For four elementary charges:
\[
q_4=4(1.60\times10^{-19})
\]
\[
q_4=6.40\times10^{-19}\ \text{C}
\]
The measured value of \(5.44\times10^{-19}\ \text{C}\) is not close to either value.
Its difference from the nearest expected value is much greater than the stated measurement uncertainty.
Therefore, simply rounding the result to three or four elementary charges is not justified.
The experimental conditions need to be investigated.
Problem 1: Brownian motion
Very small droplets experience noticeable random motion because of collisions with air molecules.
This makes it difficult to determine their terminal velocity accurately.
Improvement: Use a more suitable droplet size, take repeated observations, and improve the measurement of its average motion.
Problem 2: Limitations of the drag model
Stokes’ law assumes that the surrounding fluid behaves appropriately as a continuous medium.
For extremely small droplets, corrections may be needed.
Improvement: Apply appropriate corrections to the drag model and check the accuracy of the estimated droplet radius.
Problem 3: Measurement and calibration errors
An inaccurate electric field measurement or incorrect air viscosity could affect the calculated charge.
Improvement: Calibrate the voltage supply, verify the plate separation, and use a reliable viscosity value.
The important conclusion is that an anomalous result should be investigated rather than immediately treated as evidence against charge quantisation.
Equally, it should not simply be discarded because it does not fit the expected pattern. The reason for excluding or repeating a measurement must be justified.
Question 5
An experiment follows the same negatively charged oil droplet through four successive charge states.
During the experiment, the droplet’s radius slowly decreases. This means its resistance to motion changes.
The following measurements are recorded:
| State | Radius (\(\mu\text{m}\)) | Falling speed, \(v_f\) (mm/s) | Rising speed, \(v_r\) (mm/s) |
|---|---|---|---|
| A | 0.50 | 0.50 | 0.30 |
| B | 0.48 | 0.46 | 0.58 |
| C | 0.46 | 0.42 | 0.88 |
| D | 0.44 | 0.39 | 1.20 |
Assume:
- The electric field strength remains constant.
- The air viscosity remains constant.
- The droplet reaches terminal velocity during each measurement.
- Stokes’ law is applicable, and buoyancy can be neglected.
- Each calculated value of \(r(v_f+v_r)\) has an uncertainty of approximately \(0.006\ \mu\text{m}\,\text{mm s}^{-1}\).
The experimenter uses the relationship:
\[
|q|E=6\pi\eta r(v_f+v_r)
\]
A student compares only the rising speeds and argues that the charge changes are not quantised because the increases in rising speed are not constant.
Another student argues that the charge changes are consistent with quantisation once the changing droplet radius is considered.
a. Determine which student has the stronger argument.
b. Calculate \(r(v_f+v_r)\) for each state.
c. Determine whether the measurements are consistent with four successive integer charge states.
d. Explain why comparing rising speeds alone gives an incorrect conclusion.
e. Explain what additional information is needed to determine the numerical value of the elementary charge in coulombs.
Solution 5
The second student has the stronger argument. The measurements are consistent with successive charge states of \(4e\), \(5e\), \(6e\), and \(7e\), once the changing droplet radius is included.
The key is that the charge depends on both the droplet radius and the measured velocities.
Part a: Identify the correct approach
The governing relationship is:
\[
|q|E=6\pi\eta r(v_f+v_r)
\]
Rearranging:
\[
|q|=\frac{6\pi\eta}{E}r(v_f+v_r)
\]
Because the electric field and air viscosity remain constant:
\[
|q|\propto r(v_f+v_r)
\]
Therefore, the correct quantity to compare is \(r(v_f+v_r)\), not the rising speed alone.
Part b: Calculate the corrected values
For state A:
\[
r(v_f+v_r)=0.50(0.50+0.30)
\]
\[
=0.400
\]
For state B:
\[
r(v_f+v_r)=0.48(0.46+0.58)
\]
\[
=0.4992
\]
For state C:
\[
r(v_f+v_r)=0.46(0.42+0.88)
\]
\[
=0.598
\]
For state D:
\[
r(v_f+v_r)=0.44(0.39+1.20)
\]
\[
=0.6996
\]
The results are:
| State | \(r(v_f+v_r)\) (\(\mu\text{m}\,\text{mm s}^{-1}\)) |
|---|---|
| A | 0.400 |
| B | 0.4992 |
| C | 0.598 |
| D | 0.6996 |
Part c: Test for integer charge states
The corrected measurements are close to:
\[
0.100\times4
\]
\[
0.100\times5
\]
\[
0.100\times6
\]
\[
0.100\times7
\]
The corresponding ratios are:
\[
4.000,\quad4.992,\quad5.980,\quad6.996
\]
All are consistent with consecutive integer values within the stated uncertainty.
Therefore, the four charge states are consistent with:
\[
\boxed{4e,\quad5e,\quad6e,\quad7e}
\]
Part d: Explain the misleading prediction
The rising-speed increases are:
\[
0.58-0.30=0.28\ \text{mm/s}
\]
\[
0.88-0.58=0.30\ \text{mm/s}
\]
\[
1.20-0.88=0.32\ \text{mm/s}
\]
These increases are not identical.
A student might therefore conclude that the charge increments are unequal.
However, this ignores the decreasing droplet radius.
The drag force depends on both radius and speed:
\[
F_D=6\pi\eta rv
\]
As the radius decreases, the drag force changes even if the velocity remains the same.
Consequently, equal increments in charge do not necessarily produce equal increments in rising speed when the droplet radius is changing.
The falling-speed measurements also matter because they provide information about the effective weight of the droplet.
This is why the full relationship must be used rather than examining one measurement in isolation.
Part e: Identify the remaining limitation
The measurements establish relative charge states, but they do not independently establish the numerical value of \(e\).
To calculate the elementary charge in coulombs, the experiment must also know the calibration factor:
\[
\frac{6\pi\eta}{E}
\]
This requires reliable measurements of the air viscosity and electric field strength, together with the appropriate force model.
The distinction is important: identifying a repeated discrete pattern and accurately measuring the size of the elementary charge are related, but separate experimental achievements.
Once Millikan’s measurements established the elementary charge, physicists could combine it with Thomson’s independently measured electron charge-to-mass ratio to calculate the electron’s mass:
\[
m_e=\frac{e}{e/m_e}
\]
That is the next step in understanding why Millikan’s experiment was so important: it did not directly measure the electron’s mass, but it supplied the missing measurement needed to calculate it.