A-level Physics (Advancing Physics)/Millikan's Experiment/Worked Solutions

h = 6.63 x 10⁻³⁴ Js

c = 3 x 10⁸ ms⁻¹

g = 9.81 ms⁻²

1. Rearrange the formula above in terms of q.

${\frac {qV}{d}}=mg$

$qV=mgd$

$q={\frac {mgd}{V}}$

2. The mass of an oil drop cannot be measured easily. Express the mass of an oil drop in terms of its radius r and its density ρ, and, by substitution, find a more useful formula for q.

$\rho ={\frac {m}{\mbox{Volume}}}$

$m={\mbox{Volume}}\times \rho$

Assuming that the oil drop is spherical:

${\mbox{Volume}}={\frac {4}{3}}\pi r^{3}$

$m={\frac {4}{3}}\rho \pi r^{3}$

$q={\frac {4\pi \rho gdr^{3}}{3V}}$

3. An oil droplet of density 885kgm⁻³ and radius 1μm is held stationary in between two plates which are 10 cm apart. At what potential differences between the plates is this possible?

q must be a multiple of e, the charge on an electron:

$q=ne$, where n is an integer.

$ne={\frac {4\pi \rho gdr^{3}}{3V}}$

$V={\frac {4\pi \rho gdr^{3}}{3ne}}={\frac {4\pi \times 885\times 9.81\times 0.1\times (10^{-6})^{3}}{3\times 1.6\times 10^{-19}\times n}}={\frac {25.7}{n}}$

Then, to find some actual values, take n = 1,2,3 ... , so V = 25.7V, 12.8V, 8.6V ...

4. If the X-rays used to ionise the oil are of wavelength 1 nm, how much energy do they give to the electrons? Why does this mean that the oil drops are ionised?

$c=f\lambda$

$f={\frac {c}{\lambda }}$

$E=hf={\frac {hc}{\lambda }}={\frac {6.63\times 10^{-34}\times 3\times 10^{8}}{10^{-9}}}=1.989\times 10^{-16}{\mbox{ J}}=1.24{\mbox{ keV}}$

This provides the energy required for an electron to escape from its energy level, becoming unbound.

5. In reality, the oil drops are moving when they enter the uniform electric field. How can this be compensated for?

Either:

• Use a slightly stronger potential difference to slow the oil drop down, and then reduce the potential difference to keep the oil drop stationary.
• Measure the voltage required to keep the oil drop moving at a constant (terminal) velocity, using more sophisticated equipment.