Quantum model of light and the photoelectric effect: HSC Physics Module 7

Photon energy:

$E = h c / \lambda = (6.626 \times 10^{-34})(3.00 \times 10^8) / (3.00 \times 10^{-7}) = 6.63 \times 10^{-19}$ J $= 4.14$ eV.

Maximum kinetic energy of photoelectrons (Einstein's equation):

$K_{\max} = h f - \phi = 4.14 - 3.5 = 0.64$ eV.

Stopping voltage:

$V_s = K_{\max} / e = 0.64$ V.

Doubling the intensity: doubles the photon flux, so doubles the photocurrent (more electrons ejected per second), but does not change $K_{\max}$ or $V_s$ (each electron still absorbs only one photon of the same energy).

At $\lambda = 400$ nm: photon energy $E = 1240 / 400$ nm-eV $= 3.10$ eV, which is below the $3.5$ eV work function. No photoelectrons are emitted regardless of how bright the light is. This is the threshold-frequency phenomenon.

Markers reward correct photon energy in eV, the Einstein equation, the stopping voltage, and qualitative answers on intensity and wavelength changes consistent with the photon model.

Observation 1: threshold frequency. Below a certain frequency $f_0$, no photoelectrons are emitted no matter how bright the light. A classical wave would deliver more energy when more intense, so it should always eject electrons given enough intensity.

Einstein's explanation: light is a stream of photons, each with energy $h f$. A single photon delivers its energy to one electron all at once. If $h f < \phi$ (the work function), no electron can escape; if $h f \geq \phi$, electrons are ejected with $K_{\max} = h f - \phi$. Intensity controls photon number, not energy per photon.

Observation 2: maximum kinetic energy depends on frequency, not intensity. Brighter light of the same colour gives more photoelectrons per second but no extra energy per electron. Brighter blue light ejects faster electrons than dim blue light only in number, not in speed.

Einstein's explanation: each electron absorbs one photon and carries away energy $h f - \phi$. Doubling intensity doubles the number of absorbing electrons but each still gets exactly $h f$ of input.

Observation 3 (bonus): essentially zero delay between switching on the light and ejection of photoelectrons, even at very low intensity. A wave model would predict a build-up time; the photon model predicts instantaneous absorption.

Markers reward two observations correctly described and a clear photon-by-photon explanation linked to $h f = \phi + K_{\max}$.

Convert the photon energy using $E = hf$, then apply Einstein's equation $K_{\max} = hf - \phi$.

$E = hf = (6.626 \times 10^{-34})(7.50 \times 10^{14}) = 4.97 \times 10^{-19}\ \text{J}$.

Converting to eV: $E = \dfrac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} = 3.10\ \text{eV}$.

$K_{\max} = 3.10 - 2.20 = 0.90\ \text{eV}$.

Marks: one for correctly finding the photon energy $E = hf$ in eV, one for the correct final $K_{\max} = 0.90\ \text{eV}$ using $K_{\max} = hf - \phi$.

(a) At threshold, $K_{\max} = 0$, so all of the photon's energy just equals the work function: $f_0 = \dfrac{\phi}{h}$.

$\phi = 4.20 \times 1.602 \times 10^{-19} = 6.73 \times 10^{-19}\ \text{J}$.

$f_0 = \dfrac{6.73 \times 10^{-19}}{6.626 \times 10^{-34}} = 1.02 \times 10^{15}\ \text{Hz}$.

(b) $\lambda_0 = \dfrac{c}{f_0} = \dfrac{3.00 \times 10^8}{1.02 \times 10^{15}} = 2.95 \times 10^{-7}\ \text{m} = 295\ \text{nm}$.

Marks: one for $f_0 = \phi / h$ correctly evaluated, one for the threshold wavelength converted with the correct unit (nm or m).

(a) Photon energy

$E = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{1.80 \times 10^{-7}} = 1.104 \times 10^{-18}\ \text{J} = \dfrac{1.104 \times 10^{-18}}{1.602 \times 10^{-19}} = 6.89\ \text{eV}$.

(b) Maximum kinetic energy

$K_{\max} = E - \phi = 6.89 - 4.70 = 2.19\ \text{eV}$.

(c) Stopping voltage

Since $eV_s = K_{\max}$, and both sides are already in electron-volts, $V_s = 2.19\ \text{V}$.

Marks: one for $E = hc/\lambda$ correctly substituted and converted to eV, one for the correct value $E = 6.89\ \text{eV}$, one for $K_{\max} = hf - \phi = 2.19\ \text{eV}$, one for correctly identifying $V_s$ numerically equal to $K_{\max}$ in eV (i.e. $2.19\ \text{V}$).

(a) Gradient. $\text{gradient} = \dfrac{\Delta K_{\max}}{\Delta f} = \dfrac{(2.14 - 0.48)\ \text{eV}}{(10.0 - 6.0) \times 10^{14}\ \text{Hz}} = \dfrac{1.66\ \text{eV}}{4.0 \times 10^{14}\ \text{Hz}} = 4.15 \times 10^{-15}\ \text{eV s}$.

Converting to joules: $\text{gradient} = 4.15 \times 10^{-15} \times 1.602 \times 10^{-19} = 6.6 \times 10^{-34}\ \text{J s}$.

Since $K_{\max} = hf - \phi$ is a straight line in $f$ with gradient $h$, this value is (within rounding) Planck's constant $h = 6.626 \times 10^{-34}\ \text{J s}$.

(b) The $x$-intercept is at $f_0 \approx 4.8 \times 10^{14}\ \text{Hz}$. This is the threshold frequency: the minimum frequency of light that can eject a photoelectron from this metal ($K_{\max} = 0$ there), and $\phi = hf_0$.

Marks: one for correctly reading/using two data points, one for a numerically correct gradient with units $\text{J s}$, one for identifying the gradient as Planck's constant $h$, one for correctly reading and interpreting the $x$-intercept as the threshold frequency $f_0$.

Band-6 plan. (1) State the three observed features (threshold frequency, $K_{\max}$ depends on $f$ not intensity, no time delay). (2) For each, state the wave-model prediction and why it fails. (3) For each, give Einstein's photon explanation, naming $E = hf$ and $K_{\max} = hf - \phi$. (4) Close with an evaluative judgement on why the photon model was accepted.

Model answer. Three observations of the photoelectric effect could not be explained by the classical wave model. First, a sharp threshold frequency $f_0$ exists: below $f_0$, no photoelectrons are emitted no matter how intense or how long the light shines, whereas a classical wave should deliver energy continuously and eventually eject electrons at any frequency given enough time or intensity. Second, the maximum kinetic energy $K_{\max}$ of the ejected photoelectrons depends only on the frequency of the light, not its intensity; a wave model predicts that a more intense (higher-amplitude) wave carries more energy and so should eject faster electrons, but experiment shows intensity only changes the number of photoelectrons per second, not their energy. Third, photoelectrons are emitted with no measurable time delay, even at very low intensity, whereas a wave would need time to deliver enough energy to a single electron spread across the metal's surface.

Einstein (1905) resolved all three by proposing that light itself is quantised into photons, each carrying a fixed energy $E = hf$, and that a single photon is absorbed entirely by a single electron in one event. Since $E = hf$, a photon below the threshold ($hf < \phi$, the work function) can never free an electron regardless of how many photons arrive, explaining the threshold frequency. Because each electron absorbs exactly one photon, $K_{\max} = hf - \phi$ depends only on the frequency of that photon; increasing intensity increases the number of photons (and hence photoelectrons) per second, not the energy of each one, explaining the intensity-independence of $K_{\max}$. Because absorption is a single quantum event rather than a gradual accumulation, ejection is effectively instantaneous.

The wave model could not explain any of these because it treats energy as spread continuously across the wavefront and accumulated over time, so intensity (total wave energy) - not frequency - should determine an ejected electron's energy, and any frequency should work given enough time. The photon model succeeded because it correctly locates energy in discrete, frequency-dependent packets transferred one-for-one to electrons, which is exactly the behaviour Millikan's precise measurements of $V_s$ against $f$ later confirmed.

Marker's note: the top band addresses all three observations (not just the threshold), pairs each with both the wave-model failure and the photon-model success, and correctly states $E = hf$ and $K_{\max} = hf - \phi$ as the mechanism. A response naming only "photons carry energy" without linking to intensity-independence or the instantaneous emission caps in the middle band.

Band-6 plan. Thesis: Millikan's experiments were decisive because they turned Einstein's 1905 hypothesis into an experimentally verified law. Argue (1) what Millikan measured and how ($V_s$ vs $f$, gradient and intercept), (2) that his results matched Einstein's equation precisely despite Millikan's scepticism, (3) the broader significance - it confirmed the photon model, gave an independent measurement of $h$, and won both physicists Nobel Prizes. Weigh and conclude on "significance".

Model answer. Millikan measured the stopping voltage $V_s$ required to halt the fastest photoelectrons ejected by light of several known frequencies, using $eV_s = K_{\max} = hf - \phi$. Because $V_s$ is linear in $f$ with gradient $h/e$ and $x$-intercept $f_0 = \phi/h$, plotting $V_s$ against $f$ for a given metal gave both the work function and Planck's constant from a single graph. Millikan was initially sceptical of Einstein's "reckless" photon hypothesis and worked for nearly a decade (1907-1916) refining his apparatus - shaving metal surfaces in vacuum to remove oxide layers - specifically hoping to disprove the equation.

Instead, his results matched Einstein's equation to within about one percent, and his measured value of $h \approx 6.57 \times 10^{-34}\ \text{J s}$ agreed closely with Planck's earlier value from blackbody radiation ($6.626 \times 10^{-34}\ \text{J s}$), obtained by a completely different method. This independent agreement was powerful evidence that $h$ is a genuine universal constant, not an artefact of one experimental technique.

The significance of this was substantial. It converted the photon hypothesis from a speculative explanation of one anomaly (the photoelectric effect) into a confirmed, quantitatively precise law, directly evidencing the particle-like, quantised nature of light and reinforcing wave-particle duality. It also gave physics a second, independent route to Planck's constant, strengthening confidence that quantisation was a fundamental feature of nature rather than a mathematical convenience. Both Einstein (1921) and Millikan (1923) received Nobel Prizes substantially for this work, reflecting how central the result was to establishing quantum theory. Weighing this, Millikan's experiments were not merely confirmatory but foundational: they were the decisive experimental evidence that made the quantum model of light broadly accepted by the physics community.

Marker's note: the top band explains the method ($V_s$ vs $f$, gradient $h/e$, intercept $\phi/h$), states that Millikan's result agreed with Einstein DESPITE his intent to disprove it, links the agreement with Planck's independently-derived $h$, and reaches an explicit judgement of significance (not just a narrative of the experiment). A response that only describes the apparatus without discussing why the agreement mattered caps in the middle band.