Electronvolts to Joules: Conversion Formula and Worked Examples

A particle detector reports an electron’s kinetic energy as \(18.0\ \mathrm{keV}\), but the equation you’re using requires joules. How do you convert between the two?

The conversion factor is:

\[ \boxed{1\ \mathrm{eV}=1.602176634\times10^{-19}\ \mathrm{J}} \]

To convert electronvolts (eV) into joules (J), multiply by \(1.602176634\times10^{-19}\). To convert joules into electronvolts, divide by the same number.

For the detector reading:

\[ \begin{aligned} 18.0\ \mathrm{keV} &=18.0\times10^3\ \mathrm{eV}\\ &=18.0\times10^3\times 1.602176634\times10^{-19}\ \mathrm{J}\\ &=\boxed{2.88\times10^{-15}\ \mathrm{J}} \end{aligned} \]

The energy hasn’t changed. You’ve simply expressed it using a different unit.

Understanding where this conversion factor comes from makes it much easier to remember, apply, and use correctly in HSC Physics questions.

0101 Why does one electronvolt equal \(1.602\times10^{-19}\) joules?

Imagine an electron travelling through an evacuated tube. A power supply creates a potential difference between two metal plates, accelerating the electron from one plate towards the other.

Suppose the potential difference is \(1\ \mathrm{V}\).

How much energy does the electron gain?

The answer is \(1\ \mathrm{eV}\). This is where the electronvolt gets its name.

An electronvolt is the kinetic energy gained by an electron when it accelerates through a potential difference of one volt in a vacuum, assuming it moves in the accelerating direction.

But why does that correspond to such a small number of joules?

Start with electric potential difference

Electric potential difference describes the change in electric potential energy per unit charge.

Its defining relationship is:

\[ \Delta V=\frac{\Delta E_p}{q} \]

Rearranging:

\[ \Delta E_p=q\Delta V \]

Where:

• \(\Delta E_p\) is the change in electric potential energy, measured in joules (J).
• \(q\) is the signed electric charge, measured in coulombs (C).
• \(\Delta V\) is the electric potential difference, measured in volts (V).

One volt is equivalent to one joule per coulomb:

\[ 1\ \mathrm{V}=1\ \mathrm{\frac{J}{C}} \]

The magnitude of the elementary charge, represented by \(e\), is:

\[ e=1.602176634\times10^{-19}\ \mathrm{C} \]

This is the charge magnitude of a single proton or electron. An electron carries charge \(-e\), while a proton carries charge \(+e\).

Now consider a particle with charge magnitude \(e\) moving through a potential difference of \(1\ \mathrm{V}\) in the direction that increases its kinetic energy.

The energy gained is:

\[ \begin{aligned} E&=e\Delta V\\ &=(1.602176634\times10^{-19}\ \mathrm{C}) (1\ \mathrm{V})\\ &=(1.602176634\times10^{-19}\ \mathrm{C}) \left(1\ \mathrm{\frac{J}{C}}\right)\\ &=1.602176634\times10^{-19}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{1\ \mathrm{eV}=1.602176634\times10^{-19}\ \mathrm{J}} \]

The value of the elementary charge is exact in the modern SI system, so this conversion factor is exact too. You can find the corresponding standard value in the NIST SI Brochure.

Why use electronvolts instead of joules?

A joule is a convenient unit for everyday energy transfers, such as lifting an object or heating water.

However, the energy changes involved in individual electrons, photons, and atomic processes are incredibly small.

Writing something like \(1.602\times10^{-19}\ \mathrm{J}\) repeatedly can become cumbersome.

The electronvolt provides a more convenient scale for describing these tiny energy transfers.

Think of it this way:

• A joule is a relatively large unit of energy.
• An electronvolt is a much smaller unit of energy.
• The same physical energy therefore has a much larger numerical value when expressed in electronvolts.

0202 How to convert electronvolts into joules

Now that we know where the conversion factor comes from, the actual calculation is straightforward.

The general formula is:

\[ \boxed{E_{\mathrm{J}}= E_{\mathrm{eV}}\times1.602176634\times10^{-19}} \]

Where:

• \(E_{\mathrm{J}}\) is the energy in joules.
• \(E_{\mathrm{eV}}\) is the energy in electronvolts.

You can also write the conversion as a fraction to make the units cancel:

\[ E_{\mathrm{J}}= E_{\mathrm{eV}}\times \frac{1.602176634\times10^{-19}\ \mathrm{J}} {1\ \mathrm{eV}} \]

Notice that the electronvolt units cancel, leaving joules.

This is dimensional analysis: checking that the units on both sides of an equation are consistent.

What if the energy is given in keV or MeV?

You may encounter electronvolts with prefixes attached.

The most common are:

These prefixes work in the same way as other SI prefixes.

For example:

\[ 1\ \mathrm{keV}=1,000\ \mathrm{eV} \]

\[ 1\ \mathrm{MeV}=1,000,000\ \mathrm{eV} \]

To convert keV or MeV into joules, first account for the prefix, then apply the electronvolt-to-joule conversion.

For a refresher on powers of ten and SI prefixes, see SI Prefix Conversions for HSC Physics and Chemistry.

0303 The experiment behind the electronvolt

The electronvolt isn’t just a convenient number chosen for atomic physics. It has a direct physical connection to the work done by an electric field.

Consider how an experiment could investigate the energy gained by an electron.

Setting up the apparatus

Start with an evacuated glass tube.

At the left is a heated cathode, which releases electrons. At the right is an anode with a small opening. A high-voltage power supply connects the two electrodes, making the anode more positive than the cathode.

The electrons travel from the cathode towards the anode. They accelerate through the electric field and pass through the small opening, forming a narrow beam.

A fluorescent screen or collector at the other end allows the beam to be detected.

What should happen when the voltage increases?

Suppose the power supply creates a potential difference of \(250\ \mathrm{V}\).

If the electrons start with negligible kinetic energy, each electron should gain:

\[ \Delta K=250\ \mathrm{eV} \]

In joules:

\[ \begin{aligned} \Delta K &=250\times1.602176634\times10^{-19}\\ &=\boxed{4.01\times10^{-17}\ \mathrm{J}} \end{aligned} \]

If the accelerating voltage doubles to \(500\ \mathrm{V}\), the energy gained by each electron should also double to \(500\ \mathrm{eV}\).

The key relationship is that the energy gained is proportional to the potential difference.

Why isn’t a fluorescent screen enough?

A fluorescent screen shows where electrons arrive, but its brightness doesn’t directly tell you the kinetic energy of each electron.

Brightness also depends on how many electrons hit the screen.

We need a way to distinguish the energy of the electrons from the number of electrons in the beam.

One way to improve the experiment is to introduce a retarding potential.

A fine mesh grid is placed between the accelerating region and a collector. The grid creates an electric field that opposes the electrons’ motion. An ammeter measures the current reaching the collector.

As the retarding potential increases, fewer electrons can pass through the grid.

For an ideal beam with a narrow spread of kinetic energies, the collector current falls to zero when the retarding potential is large enough to stop the electrons.

If the electrons originally gained \(250\ \mathrm{eV}\), the stopping potential should be approximately \(250\ \mathrm{V}\).

This provides an independent way of relating kinetic energy to potential difference.

Real electron beams can have a spread of initial energies, so the current may decrease over a range of retarding voltages rather than disappearing at one perfectly sharp value.

What does this tell us?

The experiment illustrates the relationship between electric potential energy and kinetic energy.

For a particle moving under the influence of an electric field:

\[ \Delta K=-q\Delta V \]

Here:

• \(\Delta K\) is the change in kinetic energy.
• \(q\) is the signed charge of the particle.
• \(\Delta V=V_f-V_i\) is the change in electric potential.

For an electron, \(q=-e\). If it moves towards a higher electric potential, its electric potential energy decreases, and its kinetic energy increases.

This also reveals an important distinction: the electronvolt is a unit of energy, not a unit of voltage.

It can describe the energy of an electron, proton, alpha particle, or photon. The particle does not have to be an electron.

0404 Converting energy when particle charge matters

There is another detail worth understanding before tackling more complicated questions.

The electronvolt is defined using the magnitude of the elementary charge, but not every particle has a charge of magnitude \(e\).

For example:

• An electron has charge \(-e\).
• A proton has charge \(+e\).
• An alpha particle has charge \(+2e\).

A particle with twice the elementary charge gains twice as much kinetic energy when accelerated through the same potential difference.

This follows from:

\[ \Delta K=-q\Delta V \]

The mass of the particle does not appear in this equation. Mass affects the resulting speed, but not the energy transferred by the electric field.

0505 Common mistakes when converting eV into J

Before moving into practice questions, check these common errors.

A quick decision rule

When you encounter an energy conversion, identify what the question actually asks for.

For example, converting \(3.00\times10^{-19}\ \mathrm{J}\) into electronvolts gives:

\[ \begin{aligned} E&=\frac{3.00\times10^{-19}\ \mathrm{J}} {1.602176634\times10^{-19}\ \mathrm{J/eV}}\\ &=\boxed{1.87\ \mathrm{eV}} \end{aligned} \]

Notice that the units cancel in the reverse direction too.

A useful habit is to write the units at every numerical step. If your final units aren’t joules when the question asks for joules, something has gone wrong.

06Questions and solutions

The following questions progress from straightforward conversions to more demanding applications involving electric potential, particle charge, and energy conservation.

Question 1

A photon has an energy of \(3.2\ \mathrm{eV}\).

Convert its energy into joules, giving your answer to two significant figures.

Solution 1

The photon has an energy of:

\[ \boxed{5.1\times10^{-19}\ \mathrm{J}} \]

To convert electronvolts into joules, multiply by the conversion factor.

\[ 1\ \mathrm{eV}=1.602176634\times10^{-19}\ \mathrm{J} \]

Therefore:

\[ \begin{aligned} E&=3.2\ \mathrm{eV} \times1.602176634\times10^{-19} \frac{\mathrm{J}}{\mathrm{eV}}\\ &=5.127\times10^{-19}\ \mathrm{J} \end{aligned} \]

To two significant figures:

\[ \boxed{E=5.1\times10^{-19}\ \mathrm{J}} \]

The electronvolt is a unit of energy, so it can describe photons just as easily as electrons.

The conversion depends on the unit, not on the type of particle or radiation involved.

Question 2

An electron and a proton are accelerated from rest in separate runs using the same pair of parallel plates. The potential difference between the plates is \(2.4\ \mathrm{kV}\).

• The electron starts beside the negative plate and accelerates towards the positive plate.
• The proton starts beside the positive plate and accelerates towards the negative plate.

Calculate the kinetic energy gained by each particle:

(a) In electronvolts.

(b) In joules.

(c) Explain why the two particles gain the same kinetic energy despite travelling in opposite directions.

Assume both particles cross the entire potential difference and experience no collisions.

Solution 2

Both particles gain:

\[ \boxed{2.4\times10^3\ \mathrm{eV} =3.8\times10^{-16}\ \mathrm{J}} \]

Their kinetic energy gains are equal because both have a charge magnitude of \(e\), and both cross the same potential difference magnitude.

Step 1: Calculate the electron’s energy gain.

The electron has charge:

\[ q=-e \]

It travels from the negative plate to the positive plate, so its electric potential increases:

\[ \Delta V=+2.4\times10^3\ \mathrm{V} \]

Using:

\[ \Delta K=-q\Delta V \]

We obtain:

\[ \begin{aligned} \Delta K &=-(-e)(2.4\times10^3\ \mathrm{V})\\ &=2.4\times10^3\ \mathrm{eV} \end{aligned} \]

Step 2: Calculate the proton’s energy gain.

The proton has charge:

\[ q=+e \]

It travels from the positive plate to the negative plate, so:

\[ \Delta V=-2.4\times10^3\ \mathrm{V} \]

Therefore:

\[ \begin{aligned} \Delta K &=-(+e)(-2.4\times10^3\ \mathrm{V})\\ &=2.4\times10^3\ \mathrm{eV} \end{aligned} \]

Both particles gain \(2.4\ \mathrm{keV}\).

Step 3: Convert the energy into joules.

For either particle:

\[ \begin{aligned} \Delta K &=2.4\times10^3 \times1.602176634\times10^{-19}\\ &=3.845223922\times10^{-16}\ \mathrm{J} \end{aligned} \]

To two significant figures:

\[ \boxed{\Delta K=3.8\times10^{-16}\ \mathrm{J}} \]

Step 4: Explain why the energies are equal.

The electron and proton have opposite charges, but they also move through potential differences with opposite signs.

These two sign changes cancel in the kinetic energy equation.

Both particles therefore gain the same kinetic energy.

However, they do not gain the same speed. The proton has a much greater mass than the electron, so its final speed is considerably lower for the same kinetic energy.

The important lesson is that the energy gained depends on charge and potential difference, not directly on particle mass.

Question 3

An electron has an initial kinetic energy of \(500.0\ \mathrm{eV}\) at an electric potential of \(+100.0\ \mathrm{V}\).

It travels towards a detector at an electric potential of \(-150.0\ \mathrm{V}\). The detector measures its final kinetic energy as \(220.0\ \mathrm{eV}\).

Assume that any energy not accounted for by the electric field is transferred to the surroundings.

Calculate:

(a) The change in electric potential energy, in electronvolts.

(b) The energy transferred to the surroundings, in electronvolts.

(c) The energy transferred to the surroundings, in joules.

Solution 3

The electron transfers:

\[ \boxed{30.0\ \mathrm{eV} =4.81\times10^{-18}\ \mathrm{J}} \]

to the surroundings.

The important detail is that the electron moves towards a lower electric potential. Because it has negative charge, its electric potential energy increases, and its kinetic energy decreases.

Step 1: Calculate the potential difference.

Remember:

\[ \Delta V=V_f-V_i \]

Substitute:

\[ \begin{aligned} \Delta V&=-150.0-(+100.0)\\ &=-250.0\ \mathrm{V} \end{aligned} \]

Step 2: Calculate the change in electric potential energy.

For an electron:

\[ q=-e \]

Using:

\[ \Delta E_p=q\Delta V \]

We obtain:

\[ \begin{aligned} \Delta E_p &=(-e)(-250.0\ \mathrm{V})\\ &=+250.0\ \mathrm{eV} \end{aligned} \]

The electric potential energy increases by \(250.0\ \mathrm{eV}\).

In joules, this is:

\[ \begin{aligned} \Delta E_p &=250.0\times1.602176634\times10^{-19}\\ &=4.01\times10^{-17}\ \mathrm{J} \end{aligned} \]

Step 3: Calculate the observed change in kinetic energy.

The initial kinetic energy is \(500.0\ \mathrm{eV}\).

The final kinetic energy is \(220.0\ \mathrm{eV}\).

Therefore:

\[ \begin{aligned} \Delta K&=K_f-K_i\\ &=220.0-500.0\\ &=-280.0\ \mathrm{eV} \end{aligned} \]

The electron loses \(280.0\ \mathrm{eV}\) of kinetic energy.

Step 4: Determine the energy transferred to the surroundings.

The electric field accounts for an increase of \(250.0\ \mathrm{eV}\) in potential energy.

The electron’s total mechanical energy has therefore decreased by:

\[ \begin{aligned} \Delta E_{\mathrm{mechanical}} &=\Delta K+\Delta E_p\\ &=-280.0+250.0\\ &=-30.0\ \mathrm{eV} \end{aligned} \]

The missing \(30.0\ \mathrm{eV}\) has been transferred to the surroundings.

Step 5: Convert this energy into joules.

\[ \begin{aligned} E&=30.0\ \mathrm{eV} \times1.602176634\times10^{-19} \frac{\mathrm{J}}{\mathrm{eV}}\\ &=4.806529902\times10^{-18}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{E=4.81\times10^{-18}\ \mathrm{J}} \]

A tempting approach would be to assume the electron simply loses \(250.0\ \mathrm{eV}\) because it crosses a \(250.0\ \mathrm{V}\) potential difference.

That only accounts for the electric field’s contribution. The measured kinetic energy reveals an additional \(30.0\ \mathrm{eV}\) loss.

This is why energy conservation must be considered alongside unit conversion.

Question 4

A photon source emits a pulse containing \(4.0\times10^{12}\) photons.

Each photon has an energy of \(2.5\ \mathrm{eV}\). Only \(60\%\) of the photons reach a detector, which absorbs all the energy of the photons it receives.

Calculate:

(a) The energy of one photon in joules.

(b) The total energy absorbed by the detector in joules.

(c) The total energy carried by the photons that miss the detector.

Give your answers to two significant figures.

Solution 4

The energy absorbed by the detector is:

\[ \boxed{9.6\times10^{-7}\ \mathrm{J}} \]

The photons that miss the detector carry:

\[ \boxed{6.4\times10^{-7}\ \mathrm{J}} \]

We need to calculate the energy of an individual photon first, then account for how many photons reach each destination.

Step 1: Convert the energy of one photon into joules.

Each photon has an energy of \(2.5\ \mathrm{eV}\).

\[ \begin{aligned} E_{\mathrm{photon}} &=2.5\times1.602176634\times10^{-19}\\ &=4.005441585\times10^{-19}\ \mathrm{J} \end{aligned} \]

To two significant figures:

\[ \boxed{E_{\mathrm{photon}}=4.0\times10^{-19}\ \mathrm{J}} \]

Step 2: Calculate the number of photons reaching the detector.

The detector receives \(60\%\) of the original photons.

\[ \begin{aligned} N_{\mathrm{detected}} &=0.60\times4.0\times10^{12}\\ &=2.4\times10^{12}\ \mathrm{photons} \end{aligned} \]

Step 3: Calculate the total energy absorbed by the detector.

Multiply the energy per photon by the number of photons absorbed.

\[ \begin{aligned} E_{\mathrm{detector}} &=N_{\mathrm{detected}}\times E_{\mathrm{photon}}\\ &=(2.4\times10^{12}) (4.005441585\times10^{-19})\\ &=9.613059804\times10^{-7}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{E_{\mathrm{detector}}=9.6\times10^{-7}\ \mathrm{J}} \]

Step 4: Calculate the energy carried by photons that miss the detector.

The fraction that misses is:

\[ 1-0.60=0.40 \]

The number of missed photons is:

\[ \begin{aligned} N_{\mathrm{missed}} &=0.40\times4.0\times10^{12}\\ &=1.6\times10^{12} \end{aligned} \]

Their total energy is:

\[ \begin{aligned} E_{\mathrm{missed}} &=(1.6\times10^{12}) (4.005441585\times10^{-19})\\ &=6.408706536\times10^{-7}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{E_{\mathrm{missed}}=6.4\times10^{-7}\ \mathrm{J}} \]

Step 5: Check energy conservation.

The total energy emitted by the source is:

\[ \begin{aligned} E_{\mathrm{total}} &=(4.0\times10^{12}) (4.005441585\times10^{-19})\\ &=1.602176634\times10^{-6}\ \mathrm{J} \end{aligned} \]

The detector absorbs:

\[ 9.6\times10^{-7}\ \mathrm{J} \]

The missed photons carry:

\[ 6.4\times10^{-7}\ \mathrm{J} \]

Together:

\[ 9.6\times10^{-7}+6.4\times10^{-7} =1.6\times10^{-6}\ \mathrm{J} \]

The results are consistent.

The key distinction is that \(2.5\ \mathrm{eV}\) describes the energy of one photon, not the entire pulse. To find the total energy, you must account for the number of photons.

Question 5

Two particle accelerator experiments use the same potential difference of \(2.00\ \mathrm{kV}\).

Both experiments produce a particle pulse carrying a total charge magnitude of \(12.0\ \mathrm{nC}\) at the accelerator exit.

However, the particles and detector collection rates differ.

• Run A uses singly charged positive ions, each carrying charge \(+e\). The detector captures \(75.0\%\) of the particles.
• Run B uses doubly charged positive ions, each carrying charge \(+2e\). The detector captures \(100\%\) of the particles.

Assume:

• All particles start with negligible kinetic energy.
• Every particle crosses the full \(2.00\ \mathrm{kV}\) potential drop.
• Every captured particle deposits all its kinetic energy in the detector.
• No other energy losses occur.

A student argues that Run B must deposit twice as much energy as Run A because each doubly charged ion gains twice as much kinetic energy.

Determine whether this argument is correct.

Calculate:

(a) The number of particles produced in each run.

(b) The kinetic energy gained by one particle in each run, in joules.

(c) The total energy deposited in each detector, in joules.

(d) Explain why the total energy of the original particle pulses is the same, despite the different particle charges.

Solution 5

The student’s argument is incorrect.

Run B deposits:

\[ \boxed{2.4\times10^{-5}\ \mathrm{J}} \]

Run A deposits:

\[ \boxed{1.8\times10^{-5}\ \mathrm{J}} \]

Run B deposits more energy, but not twice as much. The crucial detail is that doubly charged particles carry twice the charge per particle, so a pulse with the same total charge contains only half as many particles.

We need to account for both the energy per particle and the number of particles collected.

Step 1: Calculate the number of particles in Run A.

The total charge is:

\[ Q=12.0\ \mathrm{nC}=12.0\times10^{-9}\ \mathrm{C} \]

Each particle has charge:

\[ q_A=+e=1.602176634\times10^{-19}\ \mathrm{C} \]

The number of particles is:

\[ N_A=\frac{Q}{q_A} \]

Substitute:

\[ \begin{aligned} N_A&=\frac{12.0\times10^{-9}\ \mathrm{C}} {1.602176634\times10^{-19}\ \mathrm{C}}\\ &=7.49\times10^{10}\ \mathrm{particles} \end{aligned} \]

The detector captures \(75.0\%\) of these particles:

\[ \begin{aligned} N_{A,\mathrm{detected}} &=0.750\times7.49\times10^{10}\\ &=5.62\times10^{10}\ \mathrm{particles} \end{aligned} \]

Step 2: Calculate the number of particles in Run B.

Each doubly charged ion has charge:

\[ q_B=+2e \]

Therefore:

\[ q_B=3.204353268\times10^{-19}\ \mathrm{C} \]

The number of particles is:

\[ \begin{aligned} N_B&=\frac{12.0\times10^{-9}\ \mathrm{C}} {3.204353268\times10^{-19}\ \mathrm{C}}\\ &=3.75\times10^{10}\ \mathrm{particles} \end{aligned} \]

The detector captures all of them.

Notice that Run B contains half as many particles as Run A, even though both pulses carry the same total charge.

Step 3: Calculate the energy gained by one particle in each run.

Both experiments use a potential drop of \(2.00\ \mathrm{kV}\).

For a positive particle, the kinetic energy gained is:

\[ \Delta K=q|\Delta V| \]

For Run A:

\[ \begin{aligned} \Delta K_A &=e\times2.00\times10^3\ \mathrm{V}\\ &=2.00\times10^3\ \mathrm{eV}\\ &=2.00\ \mathrm{keV} \end{aligned} \]

Convert into joules:

\[ \begin{aligned} \Delta K_A &=2.00\times10^3 \times1.602176634\times10^{-19}\\ &=3.204353268\times10^{-16}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{\Delta K_A=3.20\times10^{-16}\ \mathrm{J}} \]

For Run B:

\[ \begin{aligned} \Delta K_B &=2e\times2.00\times10^3\ \mathrm{V}\\ &=4.00\times10^3\ \mathrm{eV}\\ &=4.00\ \mathrm{keV} \end{aligned} \]

Convert into joules:

\[ \begin{aligned} \Delta K_B &=4.00\times10^3 \times1.602176634\times10^{-19}\\ &=6.408706536\times10^{-16}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{\Delta K_B=6.41\times10^{-16}\ \mathrm{J}} \]

Each particle in Run B gains twice as much energy as a particle in Run A.

However, this is only half of the problem.

Step 4: Calculate the total energy deposited in Run A.

The detector captures:

\[ N_{A,\mathrm{detected}}=5.62\times10^{10} \]

Each captured particle deposits:

\[ \Delta K_A=3.204353268\times10^{-16}\ \mathrm{J} \]

Therefore:

\[ \begin{aligned} E_A &=N_{A,\mathrm{detected}}\times\Delta K_A\\ &=(5.62\times10^{10}) (3.204353268\times10^{-16})\\ &\approx1.80\times10^{-5}\ \mathrm{J} \end{aligned} \]

Thus:

\[ \boxed{E_A=1.80\times10^{-5}\ \mathrm{J}} \]

Step 5: Calculate the total energy deposited in Run B.

All particles in Run B are captured:

\[ N_B=3.75\times10^{10} \]

Each particle deposits:

\[ \Delta K_B=6.408706536\times10^{-16}\ \mathrm{J} \]

Therefore:

\[ \begin{aligned} E_B &=N_B\times\Delta K_B\\ &=(3.75\times10^{10}) (6.408706536\times10^{-16})\\ &=2.40\times10^{-5}\ \mathrm{J} \end{aligned} \]

Thus:

\[ \boxed{E_B=2.40\times10^{-5}\ \mathrm{J}} \]

Step 6: Compare the results and explain the difference.

The ratio of the deposited energies is:

\[ \begin{aligned} \frac{E_B}{E_A} &=\frac{2.40\times10^{-5}} {1.80\times10^{-5}}\\ &=1.33 \end{aligned} \]

Run B deposits approximately \(1.33\) times as much energy as Run A.

Why isn’t the ratio 2?

Because Run B has half as many particles. Each particle gains twice as much energy, but there are only half as many particles to begin with.

In fact, before the detector collection losses, both particle pulses have exactly the same total energy.

We can see this directly from:

\[ E_{\mathrm{total}}=Q|\Delta V| \]

Substitute the total charge and potential difference:

\[ \begin{aligned} E_{\mathrm{total}} &=(12.0\times10^{-9}\ \mathrm{C}) (2.00\times10^3\ \mathrm{J/C})\\ &=2.40\times10^{-5}\ \mathrm{J} \end{aligned} \]

This result is independent of whether the beam contains singly or doubly charged particles.

The detector collection fractions then determine how much energy is actually deposited:

• Run A deposits \(75.0\%\) of the original \(2.40\times10^{-5}\ \mathrm{J}\).
• Run B deposits \(100\%\) of the original \(2.40\times10^{-5}\ \mathrm{J}\).

The subtle point is that energy per particle and total beam energy are different quantities.

A particle with twice the charge gains twice the energy across the same potential difference, but a fixed total charge contains half as many such particles.

That interaction is what changes the answer.

0706 Where to use this conversion next

Electronvolts and joules appear together in several important areas of physics.

One particularly useful connection is the photoelectric effect.

When light hits a metal surface, individual photons transfer energy to electrons. Photon energy is commonly calculated using:

\[ E=hf \]

Where:

• \(E\) is photon energy in joules.
• \(h\) is Planck’s constant.
• \(f\) is the frequency of the light.

The energy required to remove an electron from a metal is called the work function, commonly represented by \(\phi\).

It is often expressed in electronvolts.

This means you may encounter a question where photon energy is calculated in joules, but the work function is supplied in electronvolts.

Before subtracting the two energies, you must express them in the same unit.

The same principle applies to particle acceleration, atomic energy transitions, and nuclear energy calculations.

The conversion itself is simple. The more important skill is recognising what the energy represents, whether it belongs to one particle or an entire system, and whether the question requires an energy change or an absolute energy.

Once those distinctions are clear, converting between electronvolts and joules becomes a reliable part of your HSC Physics calculations.