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Questions on Millikan's oil drop experiment
Millikan's oil drop experiment found that every droplet's charge was a whole-number multiple of e = 1.6 × 10⁻¹⁹ C, the charge on the electron. A droplet held stationary between parallel plates gives QV/d = mg. Switching the field off and timing the terminal fall gives the radius through Stokes' law F = 6πηrv, and from it the mass.
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Write down the condition for a charged oil droplet to hang stationary between horizontal parallel plates, defining every symbol.
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QV/d = mg (1): Q the droplet's charge, V the pd across the plates, d their separation, m the droplet's mass. The electric force exactly supports the weight (1).State Stokes' law for the viscous force on a small sphere, and the conditions under which it applies.
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F = 6πηrv (1), for a small sphere moving slowly through a fluid of viscosity η, so that the flow around the sphere is laminar (1).State the conclusion Millikan drew from measuring the charges on hundreds of droplets.
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Every measured charge was a whole-number multiple of 1.6 × 10−19 C and never anything in between (1): electric charge is quantised, in units of the electronic charge e (1).State how the oil droplets in Millikan's experiment acquire their charge.
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By friction, as the spray breaks up on leaving the atomiser (1).In one arrangement the upper plate is at the positive potential. State and explain the sign of the charge on a droplet that can be held stationary between the plates.
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The electric force on the droplet must act upwards, to balance the weight (1). The field between the plates points downwards, away from the positive upper plate, so the droplet's charge must be negative (1).A droplet of weight 1.632 × 10−14 N hangs stationary between plates 6.0 mm apart when the pd is 612 V. Calculate its charge, and state how many electrons' worth it carries.
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Q = (weight × d)/V = 1.632 × 10−14 × 6.0 × 10−3/612 (1)
Q = 1.6 × 10−19 C (1)
Exactly one electron's worth, e itself (1)With the field off, a droplet falls at a steady 1.2 × 10−4 m s−1. Using η = 1.8 × 10−5 Pa s and oil density 880 kg m−3, calculate the droplet's radius and mass.
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At terminal speed 6πηrv = (4/3)πr3ρg (1)
r = √(9ηv/2ρg) = 1.1 × 10−6 m (1)
m = (4/3)πr3ρ (1)
m = 4.4 × 10−15 kg (1)The droplet of the previous question carries two electrons' worth of charge, 3.2 × 10−19 C. Calculate the pd needed to hold it stationary between plates 5.0 mm apart.
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V = mgd/Q (1)
V = 4.4 × 10−15 × 9.81 × 5.0 × 10−3/(3.2 × 10−19) (1)
V = 675 V (1)Explain why the falling measurement with the field switched off is an essential part of the experiment.
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The balance condition contains the droplet's mass, which is far too small to weigh directly (1). The terminal speed of the free fall, through Stokes' law, gives the radius (1); the radius gives the mass, and only then can the balance equation yield the charge (1).Millikan used oil of low volatility rather than water for his droplets. Suggest why.
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A water droplet would evaporate during the measurement (1), so its mass and radius would change while it was being timed, and the balance and terminal-speed equations would no longer describe the same droplet (1).Show that the mass of a spherical oil droplet of radius 1.2 μm is about 6.4 × 10−15 kg. Go on to calculate the electric field strength needed to hold the droplet stationary when it carries three electrons' worth of charge. Density of oil = 880 kg m−3; e = 1.60 × 10−19 C; g = 9.81 N kg−1.
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m = (4/3)πr3ρ (1) = (4/3)π × (1.2 × 10−6)3 × 880 = 6.37 × 10−15 kg, about 6.4 × 10−15 kg (1). Stationary: QE = mg, so E = mg/Q with Q = 3e = 4.80 × 10−19 C (1). E = 6.37 × 10−15 × 9.81/(4.80 × 10−19) = 1.3 × 105 V m−1 (1).A droplet carrying three electrons' worth of charge is held stationary by a pd of 440 V. Ultraviolet light then causes the droplet to lose one electron. Deduce the pd now needed to hold it stationary.
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In the balance condition QV/d = mg, the mass, g and the separation are unchanged, so V ∝ 1/Q (1). The charge falls from 3e to 2e (1), so the pd must rise by the factor 3/2: 440 × 3/2 = 660 V (1).A droplet is held stationary between the plates. The pd is then suddenly switched off. Describe and explain the droplet's subsequent motion.
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With the electric force removed the weight is unbalanced, so the droplet accelerates downwards (1). As it speeds up, the viscous drag 6πηrv grows with speed (1), until the drag equals the weight and the droplet then falls at a constant terminal speed (1).Three droplets in one run carry charges 3.2 × 10−19 C, 4.8 × 10−19 C and 8.0 × 10−19 C. Show these support charge quantisation, and state the value of e they imply.
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Dividing by 1.6 × 10−19 C gives exactly 2, 3 and 5 (1): each charge is a whole multiple of one shared unit (1) and none falls between (1). The implied quantum is e = 1.6 × 10−19 C (1).Combine Millikan's e = 1.60 × 10−19 C with Thomson's e/m = 1.76 × 1011 C kg−1 to find the mass of the electron.
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m = e ÷ (e/m) (1)
m = 1.60 × 10−19/(1.76 × 1011) (1)
m = 9.09 × 10−31 kg (1)A student's analysis of oil-drop data yields a droplet charge of 4.0 × 10−19 C. Explain why this result should send the student back to their working.
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4.0 × 10−19/(1.6 × 10−19) = 2.5, not a whole number (1). Charge only occurs in whole multiples of e, so a non-integer result indicates an experimental or arithmetic error (1).With the field off, a droplet falls at a steady 1.5 × 10−4 m s−1. With the field on, it hangs stationary between plates 6.0 mm apart carrying a pd of 755 V. Determine the number of electrons' worth of charge on the droplet. Viscosity of air = 1.8 × 10−5 Pa s; density of oil = 880 kg m−3; e = 1.60 × 10−19 C; g = 9.81 N kg−1.
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r = √(9ηv/2ρg) = 1.19 × 10−6 m (1). m = (4/3)πr3ρ = 6.15 × 10−15 kg (1). Balance: Q = mgd/V (1) = 6.15 × 10−15 × 9.81 × 6.0 × 10−3/755 = 4.8 × 10−19 C (1). Q/e = 3.0, so the droplet carries three electrons' worth of charge (1).Describe Millikan's oil drop experiment, and explain how the results establish that charge is quantised. Your answer should include the two measurements made on each droplet and the equations used.
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Charged oil droplets are sprayed between horizontal parallel plates and one is watched through a microscope (1). The pd is adjusted until the droplet hangs stationary, when the electric force balances the weight: QV/d = mg (1). With the field off, the droplet falls at terminal speed, where Stokes' drag equals the weight: 6πηrv = (4/3)πr3ρg (1). The measured terminal speed gives the radius, hence the mass, which the balance equation converts into the charge Q (1). Repeated over many droplets, every measured charge is a whole-number multiple of 1.6 × 10−19 C, with none in between (1). So charge is quantised: it exists only in integral multiples of the electronic charge e (1).
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