Blackbody Radiation and the Quantum Hypothesis
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You should be comfortable with the wave properties of light, specifically the relationship between frequency, wavelength, and the speed of light. You should also recall the definition of energy per mole and the concept of electromagnetic radiation.
Why this matters
Classical physics failed to explain the colour of hot objects, leading to the birth of quantum mechanics.
For centuries, physicists assumed that energy could be exchanged in any continuous amount. However, when studying the light emitted by heated solids, known as blackbody radiation, classical theories predicted that objects should emit infinite ultraviolet energy, a contradiction known as the ultraviolet catastrophe. Resolving this paradox required a radical new idea: that energy is not continuous, but comes in discrete packets. This single insight, proposed by Max Planck, shattered the classical worldview and laid the foundation for all modern atomic theory and quantum chemistry.
The Blackbody Problem
Imagine a perfect cavity with a small hole. If you heat this cavity, it emits light of all wavelengths. A perfect absorber and emitter is called a blackbody. In the late 19th century, scientists measured the intensity of this light at different wavelengths. They found that for each temperature, there is a specific wavelength where the intensity is maximum. As the temperature increases, this peak shifts to shorter wavelengths (bluer light). Classical wave theory, which treated light as a continuous wave, predicted that the intensity should increase indefinitely as the wavelength decreased. This meant a hot object should blast us with infinite ultraviolet radiation, which clearly does not happen. The experimental data showed a peak and then a drop-off, a shape that classical physics could not explain.
Planck’s Quantum Hypothesis
To fix the mathematical discrepancy, Max Planck proposed in 1900 that the energy of the oscillators in the cavity walls was not continuous. Instead, energy could only take on specific, discrete values. He introduced a fundamental constant, \( h \), now known as Planck’s constant. The energy \( E \) of one quantum at frequency \( \nu \) is given below; in 1905 Einstein applied the same rule to light itself, whose quanta we now call photons:
Here, \( h \approx 6.626 \times 10^{-34}\ \mathrm{J\,s} \). This means that for a given frequency, energy is quantised. You can have one quantum, two quanta, or three, but not 1.5 quanta. This discrete nature of energy exchange resolved the ultraviolet catastrophe because high-frequency modes require large energy jumps, making them statistically unlikely at lower temperatures.
Photon Energy and Frequency
The relationship \( E = h\nu \) tells us that higher frequency light carries more energy per photon. Since frequency and wavelength are inversely related (\( \nu = c/\lambda \)), we can also write:
where \( c \) is the speed of light in a vacuum, approximately \( 3.00 \times 10^8\ \mathrm{m\,s^{-1}} \). This equation is crucial for understanding why ultraviolet light can break chemical bonds while visible light often cannot. The energy of a single photon of ultraviolet light is significantly larger than that of a red photon. When we discuss atomic transitions later, we will see that the energy difference between electron orbitals must match exactly the energy of an absorbed or emitted photon.
Implications for Atomic Structure
Planck’s hypothesis was initially a mathematical trick to fit the blackbody curve. However, it implied a deep truth about the nature of matter and energy. If energy exchange is quantised, then the states of atoms must also be quantised. An electron in an atom cannot possess any arbitrary amount of energy; it can only exist in specific allowed energy levels. This concept of quantised energy levels is the cornerstone of the Bohr model and, ultimately, the quantum mechanical model of the atom. Without the quantum hypothesis, we could not explain the discrete line spectra of elements, which are the fingerprints of atomic structure.
A common misconception is that the energy of a photon depends on its intensity. In reality, intensity corresponds to the number of photons, not the energy of each individual photon. Increasing the intensity of red light increases the number of red photons, but each photon still has the same low energy. To increase the energy per photon, you must increase the frequency (shorten the wavelength). This distinction is critical for understanding photochemical reactions, where a single photon must provide enough energy to overcome an activation barrier. C
Worked example
Problem. Calculate the energy of a single photon of green light with a wavelength of \( 550\ \mathrm{nm} \). Also, determine how many such photons are required to deliver a total energy of \( 1.00\ \mathrm{J} \).
- Convert the wavelength from nanometres to metres: \( \lambda = 550\ \mathrm{nm} = 550 \times 10^{-9}\ \mathrm{m} = 5.50 \times 10^{-7}\ \mathrm{m} \).
- Use the equation \( E = \frac{hc}{\lambda} \) to find the energy of one photon. Substitute \( h = 6.626 \times 10^{-34}\ \mathrm{J\,s} \), \( c = 3.00 \times 10^8\ \mathrm{m\,s^{-1}} \), and \( \lambda = 5.50 \times 10^{-7}\ \mathrm{m} \).
- Calculate the numerator: \( hc = (6.626 \times 10^{-34}) \times (3.00 \times 10^8) = 1.9878 \times 10^{-25}\ \mathrm{J\,m} \).
- Divide by the wavelength: \( E = \frac{1.9878 \times 10^{-25}}{5.50 \times 10^{-7}} = 3.614 \times 10^{-19}\ \mathrm{J} \).
- To find the number of photons \( N \) for \( 1.00\ \mathrm{J} \), use \( N = \frac{\text{Total Energy}}{E_{\text{photon}}} \).
- Calculate \( N = \frac{1.00}{3.614 \times 10^{-19}} = 2.767 \times 10^{18} \) photons.
Units check. The energy is in Joules, and the number of photons is a dimensionless count, which is consistent with the physical quantities involved.
Three quick tests
- What is the name of the classical physics failure that predicted infinite ultraviolet emission from hot objects?
Answer
The ultraviolet catastrophe. - If the wavelength of light is halved, what happens to the energy of a single photon?
Answer
The energy doubles, because energy is inversely proportional to wavelength. - Does increasing the intensity of a light source increase the energy of each individual photon?
Answer
No, it increases the number of photons, but the energy per photon remains constant for a given frequency.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.