90-95 - Photon Model, Photoelectric Effect and Electronvolt

90-95 - Photon Model, Photoelectric Effect and Electronvolt

The photon model, the photoelectric effect, threshold frequency, work function and the electronvolt. You will use energy conservation to see why intensity, frequency and emission do different jobs.

Two models for one beam

A beam of light can behave like a wave. You have already used wave ideas for properties such as frequency, wavelength, refraction, polarisation and diffraction. Those ideas are not being thrown away.

The photoelectric effect asks a different question: when light lands on a metal surface, how is energy transferred to a single electron?

For that question, it helps to model electromagnetic radiation as photons.

Photon

A quantum (discrete packet) of electromagnetic radiation.

Photon energy

E=hf=hcλ E = hf = \frac{hc}{\lambda}

EE: energy of one photon (J); h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\mathrm{J\,s}; ff: frequency (Hz); cc: speed of light; λ\lambda: wavelength (m).

The phrase "one photon" matters. The equation does not give the total energy in a bright lamp beam. It gives the energy carried by each photon in that beam.

Historically, the wave model was very successful for many behaviours of light. The photon model developed because some observations, especially the photoelectric effect, were not explained by treating light energy as a continuous wave spread across a surface. A strong answer in this topic normally says which model explains which observation.

Misconception contrast:

Tempting ideaBetter physics
"Light is either a wave or a particle, so one model must be wrong."Different models describe different behaviours. Wave ideas describe effects such as diffraction; the photon model describes discrete energy transfer.
"A brighter beam means each photon has more energy."At a fixed frequency, brightness means more photons per second, not more energy per photon.

For an Edexcel state what is meant by command, a concise definition is enough. For explain, you need a linked reason: observation -> model -> why the model accounts for the observation.

The models developed in response to evidence. Early corpuscular accounts treated light as particles, while Huygens developed a wave account. Young’s interference experiment and diffraction supported wave behaviour; Maxwell’s electromagnetic theory connected light to oscillating electric and magnetic fields. Planck introduced quantised energy exchange, and Einstein applied light quanta to explain photoelectric observations. Later measurements of the linear KmaxK_{max}-frequency relation tested that explanation quantitatively. A successful new model must account for established observations as well as the anomaly that motivated it; the photon account therefore supplements the useful wave description rather than erasing interference and polarisation.

Photon energy and electronvolts

Photon energy is tiny on the everyday joule scale. The Edexcel data booklet gives:

h=6.63×10−34 J s1 eV=1.60×10−19 J\begin{aligned} h=6.63\times10^{-34}\,\mathrm{J\,s} \\ 1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J} \end{aligned}

Electronvolt

The energy transferred to an electron when it moves through a potential difference of 1 V: 1 eV=1.60×10−19 J1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}.

For this lesson, the useful conversion is:

Energy in eV=Energy in J1.60×10−19Energy in J=Energy in eV×1.60×10−19\begin{aligned} \text{Energy in eV}=\frac{\text{Energy in J}}{1.60\times10^{-19}} \\ \text{Energy in J}=\text{Energy in eV}\times1.60\times10^{-19} \end{aligned}

Worked example: calculate the energy of one photon of frequency 6.8×1014 Hz6.8 \times 10^{14}\,\mathrm{Hz}. Give the answer in joules and electronvolts.

Decision point 1: use E=hfE = hf because the question asks for the energy of one photon from its frequency.

E=hf=(6.63×10−34)(6.8×1014)=4.51×10−19 J\begin{aligned} E=hf=(6.63\times10^{-34})(6.8\times10^{14})=4.51\times10^{-19}\,\mathrm{J} \end{aligned}

Decision point 2: convert to eV by dividing by the joule value of 1 eV1\,\mathrm{eV}.

E=4.51×10−191.60×10−19=2.82 eV\begin{aligned} E=\frac{4.51\times10^{-19}}{1.60\times10^{-19}}=2.82\,\mathrm{eV} \end{aligned}

The powers of ten cancel cleanly because a visible/ultraviolet photon has an energy of a few electronvolts. That is the scale to expect.

Decision point 3: do not use intensity in this calculation. Intensity changes how many photons arrive each second at a given frequency. It does not change the value of hfhf for each photon.

For an Edexcel calculate command, show the equation, substitution, numerical answer and unit. If the question asks for eV, leave the answer in eV; if the next step uses 12mv2\frac{1}{2}mv^{2}, convert the energy back to joules.

The metal-surface energy budget

When a photon is absorbed by an electron in a metal, the electron may be emitted from the metal surface. An emitted electron is called a photoelectron.

The electron cannot leave for free. It needs energy to escape the metal.

Work function

The minimum energy needed to remove an electron from the surface of a metal; symbol ϕ\phi.

Threshold frequency

The minimum frequency of incident electromagnetic radiation that can cause photoelectric emission from a metal; symbol f0f_0.

At the threshold frequency:

hf0=ϕ\begin{aligned} hf_0=\phi \end{aligned}

Above the threshold frequency, any energy left after overcoming the work function becomes kinetic energy of the emitted electron. For the fastest emitted electrons:

Photoelectric equation

hf=ϕ+12mvmax⁡2hf0=ϕ hf = \phi + \frac{1}{2}mv_{\max}^{2} \qquad hf_0 = \phi

hfhf: photon energy; ϕ\phi: work function; 12mvmax⁡2\frac{1}{2}mv_{\max}^{2}: maximum kinetic energy of the photoelectrons Use consistent energy units; with mass in kg and speed in m s⁻¹, the kinetic-energy term is in joules. At the ideal threshold f=f0f=f_0, the maximum kinetic energy is zero; photoelectrons leave with positive kinetic energy for f>f0f>f_0.

The max label matters. Some electrons start deeper in the metal or lose energy before escaping, so they leave with less kinetic energy. The maximum kinetic energy comes from electrons that need only the work function to escape.

Worked example: a metal has work function 2.30 eV2.30\,\mathrm{eV}. Radiation of frequency 5.9×1014 Hz5.9 \times 10^{14}\,\mathrm{Hz} is incident on the metal. Determine whether photoelectrons are emitted. If they are, calculate the maximum kinetic energy and maximum speed of an emitted electron.

Decision point 1: compare photon energy with work function in the same unit.

E=hf=(6.63×10−34)(5.9×1014)=3.9117×10−19 JE=2.4448 eV\begin{aligned} E=hf=(6.63\times10^{-34})(5.9\times10^{14})=3.9117\times10^{-19}\,\mathrm{J} \\ E=2.4448\,\mathrm{eV} \end{aligned}

Since 2.44 eV2.44\,\mathrm{eV} is greater than 2.30 eV2.30\,\mathrm{eV}, photoelectrons can be emitted.

Decision point 2: subtract the work function to find the maximum kinetic energy.

Kmax⁡=hf−ϕ=2.4448−2.30=0.1448 eV≈0.145 eV\begin{aligned} K_{\max}=hf-\phi=2.4448-2.30=0.1448\,\mathrm{eV}\approx0.145\,\mathrm{eV} \end{aligned}

Decision point 3: convert to joules before using 12mv2\frac{1}{2}mv^{2}.

Kmax⁡=0.1448×1.60×10−19≈2.32×10−20 J\begin{aligned} K_{\max}=0.1448\times1.60\times10^{-19}\approx2.32\times10^{-20}\,\mathrm{J} \end{aligned}

Now use Kmax⁡=12mvmax⁡2K_{\max} = \frac{1}{2}mv_{\max}^{2} with the electron mass me=9.11×10−31 kgm_{e} = 9.11 \times 10^{-31}\,\mathrm{kg}.

vmax⁡=2Kmax⁡me=2(2.32×10−20)9.11×10−31≈2.3×105 m s−1\begin{aligned} v_{\max}=\sqrt{\frac{2K_{\max}}{m_e}}=\sqrt{\frac{2(2.32\times10^{-20})}{9.11\times10^{-31}}}\approx2.3\times10^5\,\mathrm{m\,s^{-1}} \end{aligned}

If the photon energy had been less than the work function, the answer would not be a negative kinetic energy. The physical answer would be: no photoelectrons are emitted.

Frequency, intensity and evidence

The photoelectric effect provides evidence for the particle nature of electromagnetic radiation because the observations make sense if energy arrives in photons.

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