Unit 20: Electrons

Millikan's Oil Drop Experiment

The Millikan oil drop experiment (1909) measured the elementary electric charge e by observing tiny, charged oil droplets suspended between two horizontal metal plates.

Experimental Setup

• A spray nozzle creates a fine mist of oil droplets that fall through a small hole into the upper chamber.
• The droplets acquire a net electric charge (usually by friction) and can be selected for observation via a microscope.
• Two parallel plates, separated by distance d, create a uniform vertical electric field E = V/d when a potential difference V is applied.
• By adjusting V, the droplet can be made to hover stationary, balancing gravitational and electric forces.

Force Balance and Derivation

For a droplet of mass m and charge q:

Gravitational force: F_g = mg

Electric force: F_e = qE

At equilibrium (F_g = F_e):

mg = qE ⟹ q = \frac{mg}{E}

Since the charge is quantized, q = ne where n is an integer. Measuring many droplets yields the smallest non‑zero value of q, identified as the elementary charge:

e = 1.602 × 10^{-19} C

Key Results and Significance

1. First direct measurement of the electron’s charge.
2. Confirmed the quantized nature of electric charge.
3. Provided a foundation for later determinations of e/m and the electron’s mass.

Motion of an Electron Beam in Electric and Magnetic Fields

When a beam of electrons (charge -e, mass m_e) enters regions with uniform electric (\(\vec{E}\)) and/or magnetic (\(\vec{B}\)) fields, its trajectory changes according to the Lorentz force:

\(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\)

Deflection by an Electric Field Alone

If only \(\vec{E}\) is present (e.g., between parallel plates), the force is constant:

\(\vec{F} = -e\vec{E}\)

Resulting acceleration: \(\vec{a} = \frac{-e\vec{E}}{m_e}\). The transverse displacement y after traveling a length L with initial speed v_x is:

\(y = \frac{1}{2} a_y t^2 = \frac{1}{2} \frac{eE}{m_e} \left(\frac{L}{v_x}\right)^2\)

Thus, deflection is proportional to E/m_e (for a given velocity).

Deflection by a Magnetic Field Alone

With only a uniform \(\vec{B}\) perpendicular to velocity, the magnetic force provides centripetal acceleration:

\(|F| = evB = \frac{m_e v^2}{r}\)

Solving for the radius of the circular path:

\(r = \frac{m_e v}{eB}\)

Hence, the trajectory is a circle whose radius depends on the momentum \(p = m_e v\) inversely on charge and field strength.

Combined Fields – Velocity Selector

When \(\vec{E}\) and \(\vec{B}\) are arranged perpendicularly (\(\vec{E}\) along y, \(\vec{B}\) along z, beam along x), the forces can cancel:

\(eE = evB ⟹ v = \frac{E}{B}\)

Only particles with this exact speed experience zero net force and pass straight through; others are deflected. This arrangement is known as a velocity selector.

Illustrative Example

Suppose an electron beam enters a region with E = 2.0 × 10^4 V/m and B = 0.10 T.

• Selected velocity: \(v = E/B = 2.0×10^4 / 0.10 = 2.0×10^5 \, m/s\).
• If the beam’s actual speed is \(1.5×10^5 \, m/s\), the net force is:
• Electric: \(F_E = eE = 1.6×10^{-19} × 2.0×10^4 = 3.2×10^{-15} N\) upward.
• Magnetic: \(F_B = evB = 1.6×10^{-19} × 1.5×10^5 × 0.10 = 2.4×10^{-15} N\) downward.
• Net: \(F_{net} = 0.8×10^{-15} N\) upward → slight upward deflection.

Thomson's Experiment for the Specific Charge (e/m)

J.J. Thomson (1897) determined the ratio of the electron’s charge to its mass (e/m_e) by measuring the deflection of cathode rays in known electric and magnetic fields.

Experimental Principle

Electrons are first accelerated through a potential difference V, gaining kinetic energy:

\(\frac{1}{2} m_e v^2 = eV ⟹ v = \sqrt{\frac{2eV}{m_e}}\)

They then enter a region where uniform \(\vec{E}\) and \(\vec{B}\) fields are applied perpendicularly (as in a velocity selector). By adjusting \(B\) until the beam is undeflected, Thomson obtained the condition:

\(eE = evB ⟹ v = E/B\)

Equating the two expressions for \(v\) yields:

\(\sqrt{\frac{2eV}{m_e}} = \frac{E}{B}\)

Squaring and solving for \(e/m_e\) gives:

\(\frac{e}{m_e} = \frac{E^2}{2VB^2}\)

Procedure and Measurements

1. Measure the accelerating voltage V (typically a few kilovolts).
2. Record the electric field strength E = V_plate/d between deflection plates.
3. Adjust the magnetic field \(B\) (using Helmholtz coils) until the electron spot returns to the undeflected position on the screen.
4. Insert the values into the formula above to compute \(e/m_e\).

Result

Thomson’s experiment gave:

\(\displaystyle \frac{e}{m_e} = 1.76 × 10^{11} \, \text{C/kg}\)

Using the later measured value of e = 1.602×10^{-19} C from Millikan, the electron mass follows as:

\(m_e = \frac{e}{e/m_e} = \frac{1.602×10^{-19}}{1.76×10^{11}} ≈ 9.10×10^{-31} \, kg\)

Historical Impact

Thomson’s measurement showed that the electron’s mass is about 1/1836 that of a hydrogen atom, establishing the electron as a subatomic constituent of all matter.

Summary of Key Formulas

Worked Numerical Problems

Problem 1: Millikan Oil Drop

An oil drop of radius \(r = 0.5 \, \mu m\) and density \(\rho = 860 \, kg/m^3\) is observed to hover when the plate voltage is \(V = 500 V\) across a separation \(d = 1.0 cm\). Find the charge on the drop.

Solution:

1. Mass: \(m = \frac{4}{3}\pi r^3 \rho = \frac{4}{3}\pi (0.5×10^{-6})^3 × 860 ≈ 4.5×10^{-13} kg\).
2. Weight: \(F_g = mg ≈ 4.5×10^{-13} × 9.8 ≈ 4.4×10^{-12} N\).
3. Electric field: \(E = V/d = 500 / 0.01 = 5.0×10^{4} V/m\).
4. Charge: \(q = F_g/E = (4.4×10^{-12})/(5.0×10^{4}) ≈ 8.8×10^{-17} C\).
5. Number of elementary charges: \(n = q/e ≈ (8.8×10^{-17})/(1.602×10^{-19}) ≈ 550\) (≈550 e).

Problem 2: Thomson’s e/m Measurement

In Thomson’s apparatus, the accelerating voltage is \(V = 2.0 kV\), the deflection plates produce \(E = 1.0×10^{4} V/m\), and the magnetic field needed for zero deflection is \(B = 2.5×10^{-3} T\). Compute \(e/m_e\).

Solution:

\(\displaystyle \frac{e}{m_e} = \frac{E^2}{2VB^2} = \frac{(1.0×10^{4})^2}{2 × (2.0×10^{3}) × (2.5×10^{-3})^2}\)

\(\displaystyle = \frac{1.0×10^{8}}{2 × 2.0×10^{3} × 6.25×10^{-6}} = \frac{1.0×10^{8}}{2.5×10^{-2}} = 4.0×10^{9} \, C/kg\).

Note: The numbers chosen for illustration give a lower value; using the accepted E, V, B from the original experiment yields \(1.76×10^{11} C/kg\).

Conclusion

The three experiments discussed—Millikan’s oil drop, Thomson’s e/m determination, and the analysis of electron beam motion in electric and magnetic fields—form the cornerstone of our understanding of the electron as a fundamental particle. They not only provide the numerical values of e and \(e/m\) but also illustrate how classical electromagnetism and mechanics combine to reveal the subatomic world.