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Background

Blackbody Radiation

The blackbody radiation is the radiation emitted by a perfect blackbody.

According to the classical physics, the energy density of the blackbody radiation is given by the Rayleigh-Jeans law:

\[d\rho(\nu, T)=\rho_\nu(T)d\nu=\frac{8\pi k_B T}{c^3}\nu^2 d\nu\]

However, this does not fully explains the experimental results, especially at high frequencies, leading to the so-called "ultraviolet catastrophe". Later, Planck proposed that the energy of the oscillators in the blackbody is quantized, leading to the Planck's law:

\[\begin{align} d\rho(\nu, T)=\rho_\nu(T)d\nu&=\frac{8\pi h}{c^3}\frac{\nu^3}{e^{\frac{h\nu}{k_B T}}-1}d\nu\\ d\rho(\lambda, T)=\rho_\lambda(T)d\lambda&=\frac{8\pi hc}{\lambda^5}\frac{d\lambda}{e^{\frac{hc}{\lambda k_B T}}-1}d\lambda \end{align}\]
Variables
  • \(\rho_\nu(T)d\nu\): radiant energy density between frequency \(\nu\) and \(\nu + d\nu\) (\(J/m^3\))
  • \(\nu\): frequency (\(s^{-1}\))
  • \(T\): temperature (\(K\))
  • \(k_B\): Boltzmann constant, the ideal gas constant over the Avogadro constant (\(1.380649 \times 10^{-23}\) \(J/K\))
  • \(h\): Planck constant (\(6.62607015 \times 10^{-34}\) \(J\cdot s\))
  • \(c\): speed of light in vacuum (\(2.99792458 \times 10^8\) \(m/s\))

The Wien's displacement law gives the wavelength \(\lambda_{max}\) at which the blackbody radiation is at maximum for a given temperature \(T\):

\[\lambda_{max}T=2.90\times 10^{-3}m\cdot K\]

In which we can differentiate the Planck's law with respect to \(\lambda\) and set it to zero to find the maximum.

\[\lambda_{max}T=\frac{hc}{4.695k_B}\]
Process

The peak wavelength is defined as \(\frac{d\rho(\lambda, T)}{d\lambda}=0\).

Let \(x=\frac{hc}{\lambda k_B T}\), we have \(\rho(\lambda, T)=\frac{8\pi k_B^5 T^5}{h^4 c^3}\frac{x^5}{e^x-1}\).

Thus, we should have \(\frac{d}{dx}(\frac{x^5}{e^x-1})=0\), or \(xe^x=5(e^x-1)\).

Solving this equation numerically gives \(x\approx 4.96511\).

Therefore, we have \(\lambda_{max}T=\frac{hc}{4.96511 k_B}\approx 2.90\times 10^{-3}m\cdot K\).

Photoelectric Effect

The photoelectric effect is the phenomenon that electrons are emitted from a material when it is exposed to light of sufficient frequency.

Energy is given in the expression of \(E=nh\nu\) and \(\Delta E=h\nu\). And for a very small package of energy emitted by light, we call it a photon, with energy \(E=h\nu\).

By conservation of energy, we have

\[\text{KE}=\frac12 mv^2=h\nu - \phi=h\nu - h\nu_0\]

where \(\phi\) is the work function of the material, representing the minimum energy required to remove an electron from the material, and \(\nu_0\) is the threshold frequency, below which no electrons are emitted.

We define electron volt (eV) as the energy gained by an electron when it is accelerated through a potential difference of one volt:

\[\begin{align} 1 eV &= (1.602 \times 10^{-19} C)(1V)\\&=1.602 \times 10^{-19} J \end{align}\]
Example

Whats the threshold frequency for sodium given its work function is \(1.82 \ eV\)?

Solution:

\(\begin{align} \phi&=1.82eV=1.82 \times 1.602 \times 10^{-19} J\\&=2.92 \times 10^{-19} J\\ \end{align}\)

\(\begin{align} \nu_0&=\frac{2.92 \times 10^{-19} J}{6.626\times 10^{-34}J\cdot s}\\&=4.40\times10^{14}\text{Hz}\\ \end{align}\)

Spectrum

The frequency \(\nu\) that corresponds to each \(n\)th energy level of the hydrogen atom is given by the Balmer formula:

\[\tilde\nu=\frac1\lambda=\frac vc=109\ 680(\frac1{2^2}-\frac1{n^2})\text{cm}^{-1}\qquad n=3,4,\dots\]

This formula can be generalized to the Rydberg formula:

\[\tilde\nu=\frac1\lambda=\frac vc=109\ 680(\frac1{n_1^2}-\frac1{n_2^2})=R_H(\frac1{n_1^2}-\frac1{n_2^2})\qquad n_2>n_1\]

Which, \(R_H=109\ 680\text{cm}^{-1}\) is the Rydberg constant for hydrogen.

Based on the relativity theory and de Broglie hypothesis, we have the de Broglie wavelength:

\[\lambda=\frac hp=\frac h{mv}\]

Angular Momentum Quantization

Linear Motion Angular Motion
mass \(m\) moment of inertia \(I\)
velocity \(v\) angular velocity \(\omega\)
linear momentum \(p=mv\) angular momentum \(l=I\omega\)
kinetic energy \(T=\frac12 mv^2\) rotational kinetic energy \(T=\frac12 I\omega^2\)

The force required to keep a particle moving in a circle is called the centripetal force:

\[f=\frac{mv^2}r=m\omega^2 r\]

For an electron moving in a circular orbit, the force is provided by the coulombic force between the electron and the nucleus, which should satisfy:

\[\frac{e^2}{4_\pi \epsilon_0 r^2}=\frac{mv^2}r\]

Which \(\epsilon_0=8.86419\times 10^{-12} C^2/(J\cdot m)\) is the permittivity of free space.

According to the classical physics, because the electron is constantly accelerating, it should constantly emit radiation and lose energy, leading to the collapse of the atom.

Bohr proposed that the angular momentum of the electron is quantized:

\[l=m_evr=n\hbar\qquad n=1,2,3,\dots\]
Variables
  • \(m_e\): mass of electron (\(9.10938356 \times 10^{-31}\) \(kg\))
  • \(\hbar\): reduced Planck constant, \(\hbar=\frac h{2\pi}\) (\(1.0545718 \times 10^{-34}\) \(J\cdot s\))

And we thus have the Bohr orbit radius:

\[r=\frac{4\pi \epsilon_0 \hbar^2 n^2}{m_e e^2}\qquad n=1,2,3,\dots\]
Example

Calculate the radius of the first Bohr orbit.

Solution:

\(\begin{align} a_0=r&=\frac{4\pi \epsilon_0 \hbar^2 n^2}{m_e e^2}\\ &=\frac{4\pi (8.86419\times 10^{-12} C^2/(J\cdot m))(1.0545718 \times 10^{-34} J\cdot s)^2 (1)^2}{(9.10938356 \times 10^{-31} kg)(1.602176634 \times 10^{-19} C)^2}\\ &= 5.29 \times 10^{-11} m= 0.529 \text{Å} \end{align}\)

The potential energy of the electron in the proton is given by:

\[V(r)=-\frac{e^2}{4\pi \epsilon_0 r}\]

Thus, the total energy of the electron is:

\[E=T+V=\frac12 m_e v^2 - \frac{e^2}{4\pi \epsilon_0 r}=-\frac{e^2}{8\pi \epsilon_0 r}\]
\[E_n=-\frac{m_e e^4}{8\epsilon_0^2 \h^2}\frac1{n^2}\qquad n=1,2,3,\dots\]

The negative sign indicates that the electron is in a bound state, which requires energy to be added to free the electron from the atom.

\[\Delta E=E_{n_2}-E_{n_1}=-\frac{m_e e^4}{8\epsilon_0^2 \h^2}(\frac1{n_2^2}-\frac1{n_1^2})=h\nu=hc\tilde\nu\]

This is called the Bohr frequency condition, which is consistent with the Rydberg formula for hydrogen spectrum.

\[\tilde\nu=\frac1\lambda=\frac vc=\frac{m_e e^4}{8\epsilon_0^2 h^3 c}(\frac1{n_1^2}-\frac1{n_2^2})\]

TO BE CONTINUED FROM chapter 1.9