Bertozzi's Experiment

How do you Measure Kinetic Energy?

When accelerating a particle with charge q from rest, the work done by an electric field with potential difference V is equal to the energy gained by the particle's kinetic store, E_k:

$q V = E_{k}$

This gives a way of measuring the kinetic energy of a particle of known charge, such as an electron

What was Bertozzi's Experiment?

Bertozzi accelerated electrons to speeds close to the speed of light and measured their kinetic energies
- The aim of the experiment was to plot the squared speeds of electrons against their kinetic energies and compare the results to Newtonian and relativistic predictions
A strong electric field was used to transfer energy to the kinetic store of the electrons
- This meant that, for an electric field of potential difference V, each electron had a kinetic energy of eV, where e is the magnitude of charge on an electron
- These electrons were fired at an aluminium target
The speed of the electrons was simply calculated using distance divided by time
- Once the electrons left the electric field, they travelled at constant speed, as no forces acted
- A signal after the electric field and a signal attached to the aluminium target were used to measure a time interval using an oscilloscope
- The distance was measured between the two signals and speed was calculated

A diagram outlining the experimental setup of Bertozzi's electron experiment

12-3-10-experimental-setup

Electrons are accelerated by an electric field and trigger a signal once they have left the field and trigger another when they hit the aluminium target, allowing speed to be calculated

The Results

With several values of speed-squared and kinetic energy, Bertozzi produced a graph showing the following results

A graph showing speed-squared against kinetic energies

12-3-10-graph-results

This graph shows the kinetic energy of the electrons, with units of m_ec², plotted against their squared speeds, with units of c². The Bertozzi results agree closely with the relativistic predictions of speeds and show electrons cannot exceed the speed of light.

This experiment's data agreed with the relativistic predictions far more than the Newtonian predictions

Verifying the Kinetic Energy of the Electrons

To avoid any doubts about whether the electrons actually did have a kinetic energy equal to eV, Bertozzi and his team measured the kinetic energy of the electrons directly as they hit the aluminium target
They did this by measuring the temperature change of the target and using the equation ΔE = mcΔθ
- This energy change is equal to the total kinetic energy lost by the incoming electrons
- The total charge transferred to the target was used to calculate the number of electrons which hit the target
With total energy transferred to the target and the number of electrons known, the team could calculate the kinetic energy per electron
- These values were in agreement with the energy transferred to the electrons by the electric field, showing the results were valid and the electrons behaved relativistically at speeds close to c

Worked example

In a repeat of the Bertozzi experiment, electrons are fired through an evacuated tube at a 0.0250 kg aluminium plate. After the experiment has run for a time, the target has accumulated −6.20 μC of charge and has increased in temperature by 1.80 °C.

Calculate the kinetic energy per electron in units of MeV.

Specific heat capacity of aluminium = 902 J kg^-1 K^-1

Answer:

Step 1: List the known quantities

Mass of aluminium, m = 0.0250 kg
Specific heat capacity of aluminium, c = 902 J kg^-1 K^-1
Change in temperature, Δθ = 1.80 °C
Charge transferred to aluminium plate, Q = −6.20 μC
Charge of an electron, e = −1.60 × 10⁻¹⁹ C

Step 2: Write out the relevant equation

Thermal energy transferred to aluminium, ΔE = mcΔθ

Step 3: Calculate the energy transferred to the aluminium's thermal store

The thermal energy gained by the plate is:

$Δ E = m c Δ θ = 0.0250 \times 902 \times 1.80 = 40.59 J$

Step 4: Calculate the number of electrons incident on the aluminium plate

Each electron transfers a charge of e to the plate
The total charge transferred to the plate divided by e will give the number of electrons, n, incident on the plate:

$n = \frac{Q}{e} = \frac{- 6.20 \times 10^{- 6}}{- 1.60 \times 10^{- 19}} = 3.875 \times 10^{13}$

Step 5: Calculate the kinetic energy per electron

The total thermal energy gained by the plate is equal to the total kinetic energy, E_k,tot , of all the electrons hitting it
E_k,totdivided by the number of electrons gives the kinetic energy per electron, E_k :

$E_{k} = \frac{E_{k, t o t}}{n} = \frac{Δ E}{n} = \frac{40.59}{3.875 \times 10^{13}} = 1.0475 \times 10^{- 12} J$

Step 6: Convert the kinetic energy per electron into mega-electronvolts

Recall that 1 eV is equal to 1.60 × 10^-19 J:

$E_{k} = \frac{1.0475 \times 10^{- 12}}{1.60 \times 10^{- 19}} = 6.547 \times 10^{6} eV$

Finally, the answer must be to 3 significant figures, as all the figures in the question were given to this number:

$E_{k} = 6.55 MeV$

Examiner Tip

You may be asked on either method of determining the kinetic energy of each electron so expect to see the thermal energy equation and work done by an electric field in this topic.

Bertozzi's Experiment (AQA A Level Physics)

Revision Note