Photon Energy and Electronvolts | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Frequency sets the energy of one photon

Frequency is the number of oscillations of the electromagnetic wave each second. One hertz means one oscillation each second. The frequency determines the energy carried by each photon.

E=hf

Symbols

$E$ = energy of one photon (J)
$h$ = Planck constant, 6.63 × 10⁻³⁴ (J s)
$f$ = frequency of the radiation (Hz)

Planck's constant does not change.
A higher frequency gives more energy to each photon.
If the frequency doubles, the energy of each photon doubles.

Worked example

Find the energy of one photon

Calculate the energy of one photon with a frequency of 5.0 × 10¹⁴ Hz.

E=hf

\begin{aligned} E&=(6.63\times10^{-34})\\ &\quad\times(5.0\times10^{14}) \end{aligned}

E=3.3\times10^{-19}\ \mathrm{J}

Answer

E=3.3\times10^{-19}\ \mathrm{J}

A question may give wavelength instead of frequency. From AS Waves, electromagnetic radiation in a vacuum obeys the relation below.

c=f\lambda

Symbols

$c$ = speed of electromagnetic radiation in a vacuum (m s⁻¹)
$f$ = frequency of the radiation (Hz)
$\lambda$ = wavelength of the radiation (m)

The speed in a vacuum is fixed. Therefore, when frequency increases, wavelength decreases.

Replace frequency in the photon-energy equation by speed divided by wavelength.

E=\dfrac{hc}{\lambda}

Your turnconcept check

Radiation A has twice the frequency of radiation B. Compare the energy of one photon from each radiation.

Show worked answer

One photon from A has twice the energy of one photon from B. The frequency doubles while Planck's constant remains fixed.

Higher frequency means more energy per photon. Shorter wavelength also means more energy per photon.

Use electronvolts for very small energies

The energy of one photon is often much smaller than one joule. Writing these values in joules requires a negative power of ten. The is a smaller unit that is convenient for particle and photon energies.

One electronvolt is the energy transferred when an electron moves through a potential difference of 1 V. The conversion comes from the energy transferred to a charge moving through a potential difference.

\Delta E=QV

\Delta E=(1.60\times10^{-19}\ \mathrm{C})(1\ \mathrm{V})

One volt is one joule per coulomb, so the coulomb units cancel. The result is the energy called one electronvolt:

1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}

Changing between joules and electronvolts changes only the unit, not the physical energy.
To convert from eV to J, multiply by 1.60 × 10⁻¹⁹.
To convert from J to eV, divide by 1.60 × 10⁻¹⁹.

Worked example

Convert electronvolts to joules

Convert a photon energy of 4.0 eV to joules.

E=(4.0)(1.60\times10^{-19})

E=6.4\times10^{-19}\ \mathrm{J}

Answer

E=6.4\times10^{-19}\ \mathrm{J}

Your turnunit retrieval check

Express a photon energy of 6.4 × 10⁻¹⁹ J in electronvolts.

Show worked answer

Divide by 1.60 × 10⁻¹⁹ J per eV. The photon energy is 4.0 eV.

Frequency sets the energy of one photon

Frequency is the number of oscillations of the electromagnetic wave each second. One hertz means one oscillation each second. The frequency determines the energy carried by each photon.

E=hf

Symbols

$E$ = energy of one photon (J)
$h$ = Planck constant, 6.63 × 10⁻³⁴ (J s)
$f$ = frequency of the radiation (Hz)

Planck's constant does not change.
A higher frequency gives more energy to each photon.
If the frequency doubles, the energy of each photon doubles.

Worked example

Find the energy of one photon

Calculate the energy of one photon with a frequency of 5.0 × 10¹⁴ Hz.

E=hf

\begin{aligned} E&=(6.63\times10^{-34})\\ &\quad\times(5.0\times10^{14}) \end{aligned}

E=3.3\times10^{-19}\ \mathrm{J}

Answer

E=3.3\times10^{-19}\ \mathrm{J}

A question may give wavelength instead of frequency. From AS Waves, electromagnetic radiation in a vacuum obeys the relation below.

c=f\lambda

Symbols

$c$ = speed of electromagnetic radiation in a vacuum (m s⁻¹)
$f$ = frequency of the radiation (Hz)
$\lambda$ = wavelength of the radiation (m)

The speed in a vacuum is fixed. Therefore, when frequency increases, wavelength decreases.

Replace frequency in the photon-energy equation by speed divided by wavelength.

E=\dfrac{hc}{\lambda}

Your turnconcept check

Radiation A has twice the frequency of radiation B. Compare the energy of one photon from each radiation.

Show worked answer

One photon from A has twice the energy of one photon from B. The frequency doubles while Planck's constant remains fixed.

Higher frequency means more energy per photon. Shorter wavelength also means more energy per photon.

Use electronvolts for very small energies

\Delta E=QV

\Delta E=(1.60\times10^{-19}\ \mathrm{C})(1\ \mathrm{V})

One volt is one joule per coulomb, so the coulomb units cancel. The result is the energy called one electronvolt:

1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}

Changing between joules and electronvolts changes only the unit, not the physical energy.
To convert from eV to J, multiply by 1.60 × 10⁻¹⁹.
To convert from J to eV, divide by 1.60 × 10⁻¹⁹.

Worked example

Convert electronvolts to joules

Convert a photon energy of 4.0 eV to joules.

E=(4.0)(1.60\times10^{-19})

E=6.4\times10^{-19}\ \mathrm{J}

Answer

E=6.4\times10^{-19}\ \mathrm{J}

Your turnunit retrieval check

Express a photon energy of 6.4 × 10⁻¹⁹ J in electronvolts.

Show worked answer

Divide by 1.60 × 10⁻¹⁹ J per eV. The photon energy is 4.0 eV.