Mass and Energy, Particles and Waves

Mass-Energy Equivalence Principle

Mass and energy are identical and can be converted into each other.

\[E=mc^2\]

where $c$ is the speed of light, $2.9979 \times 10^{10}\ \text{cm/sec}$.

Electron Volt (eV)

Electron volt (eV): The kinetic energy gained by an electron passing through a potential difference of 1 volt

\[\begin{align*} 1 \text{eV} &= 1.60219 \times 10^{-19}\ \text{C}\cdot \text{V} \\ &= 1.60219 \times 10^{-19}\ \text{J} \end{align*}\]

Mass and Energy of Moving Objects

According to the theory of relativity, the mass of a moving object increases relative to an observer, and the equation relating the speed and mass of a moving object is defined as follows:

\[m=\frac {m_0}{\sqrt{1-v^2/c^2}} \tag{1}\]

$m_0$: rest mass, $v$: speed

The total energy of a particle is the sum of its rest-mass energy and kinetic energy, so the following holds:

\[E_{\text{total}} = E_{\text{rest}}+E_{\text{kinetic}} = mc^2\] \[\begin{align*} E_{\text{kinetic}} &= E_{\text{total}}-E_{\text{rest}} \\ &= mc^2 - m_0c^2 \\ &= m_0c^2\left[\frac {1}{\sqrt{1-v^2/c^2}} - 1\right] \tag{2} \end{align*}\]

Particularly when $v\ll c$, if we set $\cfrac{v^2}{c^2} = \epsilon$ and approximate using Taylor expansion around $\epsilon = 0$ (i.e., Maclaurin expansion):

\[\begin{align*} E_{\text{kinetic}} &= m_0c^2\left[\frac {1}{\sqrt{1-\epsilon}} - 1\right] \\ &= m_0c^2\left[ (1-\epsilon)^{-\frac{1}{2}} - 1 \right] \\ &= m_0c^2\left[ \left( 1 + \frac{1}{2}\epsilon + O(\epsilon^2) \right) - 1 \right] \\ &\approx m_0c^2\left[ \left( 1 + \frac{1}{2}\epsilon \right) - 1 \right] \\ &= \frac{1}{2}m_0c^2\epsilon \\ &= \frac {1}{2}m_0v^2 \tag{3} \end{align*}\]

This becomes the same as the kinetic energy formula in classical mechanics. Practically, when $v\leq 0.2c$ or $E_{\text{kinetic}} \leq 0.02E_{\text{rest}}$, we can consider $v\ll c$ and use this approximation (i.e., ignore relativistic effects) to obtain sufficiently accurate values.

Electrons

Since the rest-mass energy of an electron is $E_{\text{rest}}=m_ec^2=0.511 \text{MeV}$, the relativistic kinetic energy formula should be applied when the electron’s kinetic energy exceeds $0.02\times 0.511 \text{MeV}=0.010 \text{MeV}=10 \text{keV}$. In nuclear engineering, the energy of electrons often exceeds 10 keV, so equation (2) must be applied in most cases.

Neutrons

The rest-mass energy of a neutron is approximately 1000 MeV, so $0.02E_{rest}=20\text{MeV}$. Since it is rare to deal with neutron kinetic energies exceeding 20 MeV in nuclear engineering, equation (3) is typically used to calculate neutron kinetic energy.

Photons

Equations (2) and (3) are valid only when the rest mass is not zero, so they cannot be applied to photons with zero rest mass. The total energy of a photon is calculated using the following equation:

\[E = h\nu \tag{4}\]

$h$: Planck’s constant ($4.316 \times 10^{-15} \text{eV}\cdot\text{s}$), $\nu$: frequency of the electromagnetic wave

Matter Waves

Everything in nature is both a particle and a wave simultaneously. That is, all particles have a corresponding wavelength (de Broglie wavelength). The wavelength $\lambda$ is a function of momentum $p$ and Planck’s constant $h$.

\[\lambda = \frac {h}{p} \tag{5}\]

Also, momentum $p$ is defined by the following equation:

\[p = mv \tag{6}\]

Neglecting Relativistic Effects (e.g., Neutrons)

Since kinetic energy is $E=1/2 mv^2$, expressing equation (6) as a function of energy gives:

\[p=\sqrt{2mE} \tag{7}\]

Substituting this into equation (5), the particle’s wavelength becomes:

\[\lambda = \frac {h}{\sqrt{2mE}} \tag{8}\]

This equation is applied when calculating the de Broglie wavelength of neutrons in nuclear engineering. Substituting the rest mass of a neutron gives:

\[\lambda = \frac {2.860 \times 10^{-9}}{\sqrt{E}} \tag{9}\]

where $\lambda$ is in cm and $E$ is the neutron’s kinetic energy expressed in eV.

Considering Relativistic Effects (e.g., Electrons)

Calculate momentum $p$ by directly solving the relativistic equations:

\[p=\frac {1}{c} \sqrt{E^2_{\text{total}}-E^2_{\text{rest}}} \tag{10}\]

Then the de Broglie wavelength is:

\[\lambda = \frac {hc}{\sqrt{E_{\text{total}}-E_{\text{rest}}}} \tag{11}\]

Particles with Zero Rest Mass (e.g., Photons)

For particles with zero rest mass, momentum cannot be calculated using equation (6), so it is expressed as:

\[p=\frac {E}{c} \tag{12}\]

Substituting equation (12) into equation (5):

\[\lambda = \frac {hc}{E} \tag{13}\]

Substituting the values of $h$ and $c$, the final equation for wavelength is:

\[\lambda = \frac {1.240 \times 10^{-6}}{E} \tag{14}\]

where $\lambda$ is in meters and $E$ is in eV.