Quantum Physics

A blackbody is an ideal absorber and emitter of radiation. Classical physics predicted the Rayleigh-Jeans Law:

$I(\lambda, T) = \frac{2\pi c k_B T}{\lambda^4}$

This works at long wavelengths but fails at short wavelengths, predicting infinite energy—a problem known as the ultraviolet catastrophe.

Planck resolved this by proposing quantized energy emission:

$E_n = n h f, \quad (n = 1, 2, 3, \ldots)$

This led to Planck's Law:

$I(\lambda, T) = \frac{2\pi h c^2}{\lambda^5} \left( \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1} \right)$

This law matched experimental data and introduced energy quantization.

Classical theory predicted energy of photoelectrons depended on intensity, not frequency. But experiments showed:

1. Emission is instantaneous Electrons are emitted from the metal surface immediately after exposure to light, regardless of intensity, contradicting classical predictions of delayed emission.

2. There's a minimum threshold frequency Photoemission only occurs if the incident light has a frequency above a certain threshold. Below this frequency, no electrons are emitted, regardless of intensity.

3. Electron energy increases with frequency, not intensity The kinetic energy of emitted electrons increases with the frequency of the light, not its brightness. This means higher-frequency photons transfer more energy to the electrons.

Einstein proposed light consists of photons with energy:

$E = h f$

The kinetic energy of emitted electrons is:

$K_{\text{max}} = h f - \phi_0$

Photoelectric Experiment Setup

A vacuum tube with a photoemissive cathode and anode, connected to a circuit with a variable voltage and ammeter. Light is directed on the cathode.

• More intensity → more photocurrent (above threshold)
• Stopping potential $V_0$ satisfies: $eV_0 = h f - \phi_0$
• Graph of $V_0$ vs $f$ is linear, slope = $h/e$

Laws of Photoelectric Emission

1. Threshold frequency $f_0$ below which no emission occurs
2. Photocurrent ∝ intensity (above threshold)
3. $K_{\text{max}} \propto f$, independent of intensity
4. Emission occurs without delay

Photon Theory vs Classical Theory

Classical theory fails to explain instantaneous emission, threshold frequency, and intensity-independent kinetic energy. Quantum theory, with photon model, explains all.

Work Function Table

The photoelectric effect supports the quantum nature of light and matter, illustrating the duality central to quantum mechanics.

In the early 20th century, Albert Einstein's photon theory (1905) introduced the idea that electromagnetic radiation exhibits particle-like properties. Expanding on this concept, in 1924, French physicist Louis de Broglie proposed that not only light but all matter possesses wave-like characteristics. This was a profound departure from classical mechanics and gave rise to the principle of wave-particle duality, a cornerstone of quantum mechanics.

De Broglie hypothesized that a particle of mass $m$ moving with velocity $v$ has an associated wavelength $\lambda$, known as the de Broglie wavelength, given by:

$\lambda = \frac{h}{p} = \frac{h}{mv}$

where:

• $\lambda$ is the de Broglie wavelength,
• $h = 6.626 \times 10^{-34} \, \text{Js}$ is Planck's constant,
• $p = mv$ is the linear momentum of the particle.

This revolutionary idea was experimentally verified in 1927 by Davisson and Germer, who demonstrated that electrons, when diffracted by a nickel crystal, produced interference patterns characteristic of waves. Their findings provided conclusive evidence that particles such as electrons exhibit wave-like behavior.

Relationship Between Energy and Wavelength

From Einstein's theory of relativity:

E = mc^2 \tag{4.8}

From Planck's quantum theory:

E = hf = \frac{hc}{\lambda} \tag{4.9}

Equating the two expressions for energy gives:

\lambda = \frac{hc}{mc^2} \tag{4.10}

This equation is valid only for particles moving at the speed of light, such as photons. However, for massive particles like electrons that travel at speeds less than $c$, the general de Broglie expression becomes:

\lambda = \frac{h}{mv} = \frac{h}{p} \tag{4.11}

The idea that matter exhibits both wave-like and particle-like properties is referred to as wave-particle duality. This concept underlies the theoretical framework of quantum mechanics, also known as wave mechanics.

Example 5

Calculate the de Broglie wavelength of:

a. An electron accelerated through a potential difference $V = 100 \, \text{V}$

b. A proton accelerated through the same potential

Given:

• $h = 6.626 \times 10^{-34} \, \text{Js}$
• $e = 1.602 \times 10^{-19} \, \text{C}$
• $m_e = 9.11 \times 10^{-31} \, \text{kg}$
• $m_p = 1.673 \times 10^{-27} \, \text{kg}$

Solution:

The kinetic energy gained by a charged particle is:

$eV = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2eV}{m}}$

The de Broglie wavelength becomes:

$\lambda = \frac{h}{\sqrt{2meV}}$

a. For an electron:

$\lambda_e = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 9.11 \times 10^{-31} \cdot 1.602 \times 10^{-19} \cdot 100}} \approx 1.23 \times 10^{-10} \, \text{m} = 0.123 \text{nm}$

b. For a proton:

$\lambda_p = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 1.673 \times 10^{-27} \cdot 1.602 \times 10^{-19} \cdot 100}} \approx 0.0029 \text{nm}$

Example 6

Compare the de Broglie wavelength of:

a. A bullet of mass $100 \, \text{g} = 0.1 \, \text{kg}$ traveling at $500 \, \text{m/s}$

b. An electron traveling at the same speed

Solution:

a For the bullet:

$\lambda = \frac{6.626 \times 10^{-34}}{0.1 \cdot 500} = 1.3252 \times 10^{-35} \, \text{m}$

b. For the electron:

$\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 500} \approx 1.45 \times 10^{-6} \, \text{m}$

Interpretation:

The electron has a significantly larger wavelength than the bullet, making its wave properties observable, unlike macroscopic objects whose wavelengths are practically undetectable.

Example 7

Compute the de Broglie wavelength of:

a. An electron

b. A neutron

Each traveling at: $v = 1.0 \times 10^5 \, \text{m/s}$

Given:

• $m_e = 9.11 \times 10^{-31} \, \text{kg}$
• $m_n = 1.675 \times 10^{-27} \, \text{kg}$

Solution:

a. Electron:

$\lambda_e = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 1.0 \times 10^5} \approx 7.27 \times 10^{-9} \, \text{m} = 7.27 \, \text{nm}$

b. Neutron:

$\lambda_n = \frac{6.626 \times 10^{-34}}{1.675 \times 10^{-27} \cdot 1.0 \times 10^5} \approx 3.95 \times 10^{-12} \, \text{m} = 3.95 \, \text{pm}$

Diagram: X-Ray Tube

The production of X-rays involves the acceleration of electrons through a high potential difference towards a metal target (typically tungsten). When these high-energy electrons strike the target, X-rays are emitted.

Components:

• Heated filament (cathode) emits electrons
• Electrons accelerated towards the anode
• X-rays are emitted upon striking the metal target

• The de Broglie hypothesis unifies wave and particle properties.
• Explains electron diffraction and supports technologies like electron microscopes.
• Wave nature is observable mainly in microscopic particles due to their small mass and high speeds.