Question

The electron-volt, \(\mathrm{eV}\), is a unit of energy \(\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~J}, 1 \mathrm{MeV}=\right.\) \(1.602 \cdot 10^{-13} \mathrm{~J}\) ). Since the unit of momentum is an energy unit divided by a velocity unit, nuclear physicists usually specify momenta of nuclei in units of \(\mathrm{MeV} / \mathrm{c},\) where \(c\) is the speed of light \(\left(c=2.998 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\right) .\) In the same units, the mass of a proton \(\left(1.673 \cdot 10^{-27} \mathrm{~kg}\right)\) is given as \(938.3 \mathrm{MeV} / \mathrm{c}^{2}\). If a proton moves with a speed of \(17,400 \mathrm{~km} / \mathrm{s}\), what is its momentum in units of \(\mathrm{MeV} / \mathrm{c}\) ?

Short answer

Answer: The momentum of the proton is approximately 0.999 MeV/c.

Step-by-step solution

Convert the given speed to meters per second (m/s)

To convert the given speed from kilometers per second (km/s) to meters per second (m/s), we use the following conversion factor: 1 km = 1000 m. So, we have: \((17,400 \, \mathrm{km/s})(1,000 \, \mathrm{m/km}) = 17,400,000 \, \mathrm{m/s}\).

Calculate the momentum of the proton

The momentum (p) of an object can be calculated using the formula: \(p = m \cdot v\), where \(m\) is the mass of the object and \(v\) is its velocity. We already know the mass of the proton (\(1.673 \cdot 10^{-27} \, \mathrm{kg}\)) and its speed (\(17,400,000 \, \mathrm{m/s}\)) from the previous step, so we can calculate its momentum as follows: \(p = (1.673 \cdot 10^{-27} \, \mathrm{kg}) (17,400,000 \, \mathrm{m/s}) = 2.9105 \cdot 10^{-20} \, \mathrm{kg \cdot m/s}\).

Convert the momentum to units of MeV/c

To convert the momentum to units of MeV/c, we'll use the given conversion factors. First, we'll convert the momentum to J m/s, using the energy conversion factor \(1 \, \mathrm{eV} = 1.602 \cdot 10^{-19} \, \mathrm{J}\). Then, we'll determine the proportionality constant needed for converting the momentum to MeV/c. Finally, we'll divide the momentum in J m/s by the proportionality constant to obtain the momentum in units of MeV/c. Proportionality constant = \(\frac{1.673 \cdot 10^{-27} \, \mathrm{kg} \cdot c}{938.3 \, \mathrm{MeV}/c^{2}} = 2.9107299 \cdot 10^{-20} \, \mathrm{J \cdot m/s}\) Now, divide the momentum in J m/s by the proportionality constant: \(\frac{2.9105 \cdot 10^{-20} \, \mathrm{J \cdot m/s}}{2.9107299 \cdot 10^{-20} \, \mathrm{J \cdot m/s}} = 0.999 \, \mathrm{MeV/c}\) (rounded to 3 decimal places). So, the momentum of the proton is approximately \(0.999 \, \mathrm{MeV/c}\).

Key concepts

Nuclear Physics and Electron-Volt

Nuclear physics is the branch of physics that studies the constituents and interactions of atomic nuclei. One common unit of energy used in nuclear physics is the electron-volt (eV), which is the amount of kinetic energy gained or lost by an electron when it moves through an electric potential difference of one volt. In more tangible terms, one electron-volt is equal to approximately \(1.602 \cdot 10^{-19} \(\mathrm{J}\)) or for higher energies, one Mega-electron-volt (MeV) is \(1.602 \cdot 10^{-13} \(\mathrm{J}\)).

Understanding Momentum Calculation

Momentum is a fundamental concept in physics representing the quantity of motion an object has. For a proton or any particle, the momentum (\(p\)) is calculated by multiplying its mass (\(m\)) by its velocity (\(v\)): \(p = m \cdot v\). In nuclear physics, it's often more convenient to use the units of MeV/c, where \(c\) represents the speed of light, to express momentum. This aligns with the relativistic nature of particles moving at high speeds, where classical calculations of momentum and energy intertwine.

When calculating momentum in this way, remember to convert measures into the consistent set of units used in the field, which often involves converting from kilograms or meters per second to MeV/c. This allows nuclear physicists to work within a framework that's compatible with both relativistic effects and quantum mechanics.

Proton Mass and Energy Equivalence

The mass-energy equivalence principle, conceptualized by Einstein's famous equation \(E=mc^2\), reveals that mass can be considered a form of energy. In the context of nuclear physics, the mass of a proton is often expressed as \(938.3 \mathrm{MeV/c^2}\).

This corresponds to a rest energy that a proton would have purely based on its mass. When we talk about protons moving at a considerable fraction of the speed of light, as in high-energy physics experiments or naturally occurring in cosmic rays, their kinetic energy - and thus their total energy (mass-energy and kinetic energy) - significantly increases, which can also increase the resulting momentum as observed in various physical phenomena and experiments.

For students mastering nuclear physics, understanding the mass-energy equivalence not only helps with momentum calculations but also provides a deeper insight into the behavior of particles under high-energy conditions.