Problem 22 - University Physics with Modern Physics

Question

The electron-volt, \(\mathrm{eV},\) is a unit of energy \((1 \mathrm{eV}=\) \(\left.1.602 \cdot 10^{-19} \mathrm{~J}, 1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right) .\) Since the unit of \(\mathrm{mo}\) mentum is an energy unit divided by a velocity unit, nuclear physicists usually specify momenta of nuclei in units of \(\mathrm{MeV} / c,\) where \(c\) is the speed of light \(\left(c=2.998 \cdot 10^{9} \mathrm{~m} / \mathrm{s}\right) .\) 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 \(\mathrm{MeV} / \mathrm{c}\) ?

Step-by-Step Solution

Answer

Answer: The momentum of the proton is approximately 6.015 MeV/c.

Step 1: Converting units of speed

First, we need to convert the speed of the proton from km/s to m/s. We know that 1 km = 1000 m, so 17,400 km/s can be converted to m/s as follows: \[17,400\,km/s \times \frac{1000\,m}{1\,km} = 17,400 \times 10^3\,m/s\]

Step 2: Finding the momentum in SI units

Next, we can find the momentum of the proton in SI units by using the formula: \[p = mv\] where p is the momentum, m is the mass of the proton (1.673 * 10^-27 kg), and v is the speed we found in Step 1. \[p = (1.673 \times 10^{-27}\,kg)(17,400 \times 10^3\,m/s) = 2.90862 \times 10^{-23}\,kg\cdot m/s\]

Step 3: Converting the momentum to MeV/c

Now, we convert the momentum to MeV/c units. We have to divide by two conversion factors: The energy conversion factor (1.602 * 10^-13 J/MeV) and the speed of light (2.998 * 10⁹ m/s). So we can calculate the momentum like this: \[p[\frac{MeV}{c}] = \frac{p_{(SI\,units)}\, [\frac{kg\,m}{s}]}{ENERGY\,CONVERSION\,FACTOR\,[\frac{J}{MeV}] \times c\,[\frac{m}{s}]}\] \[p[\frac{MeV}{c}] = \frac{2.90862 \times 10^{-23}\,\frac{kg\cdot m}{s}}{(1.602 \times 10^{-13}\,\frac{J}{MeV}) \times (2.998 \times 10^9\,\frac{m}{s})} = 6.015 \mathrm{MeV}/\mathrm{c}\] So, the momentum of the proton in units of MeV/c is approximately 6.015 MeV/c.

Key Concepts

Proton

A proton is one of the basic constituents of matter. It's a subatomic particle found in the nucleus of an atom and carries a positive electric charge of +1. Protons, along with neutrons, make up the atomic nucleus.
The properties of the proton such as its mass are crucial in physics and chemistry. It has a mass of approximately \(1.673 \times 10^{-27} \text{ kg}\), which is relatively about 1836 times the mass of an electron.
Knowing the mass and charge of the proton, scientists can calculate its momentum when it is moving at a certain speed. Momentum plays a vital role in understanding how the proton will behave when subjected to external forces.

Electron-Volt

The electron-volt (eV) is a commonly used unit of energy in the field of particle physics. It provides a measure of energy within the context of atomic and subatomic processes.
Specifically, 1 eV represents the energy gained by an electron when it is accelerated through an electric potential difference of 1 volt. This is equivalently \(1.602 \times 10^{-19} \text{ Joules}\).
Because particle interactions often deal with tiny energies, the electron-volt is a convenient unit, especially expressed in MeV (million electron-volts) or GeV (billion electron-volts) for higher scales.

Energy Conversion

Energy conversion is significant when working with different units, such as Joules and electron-volts, which are critical in equations involving particle momentum.
In physics, understanding how to convert energy is vital for calculating and expressing measurements appropriately. For instance, 1 MeV \(= 1.602 \times 10^{-13} \text{ Joules}\).
When calculating momentum, we often start with SI units and need to convert to units like \(\text{MeV}/c\), where \(c\) denotes the speed of light, allowing for practical communication in nuclear and particle physics.

Speed of Light

The speed of light, denoted as \(c\), is fundamental to the theories of physics, especially in the realms of relativity and quantum mechanics.
It is a constant value \(c = 2.998 \times 10^8 \text{ m/s}\), which serves as the ultimate speed limit for the transfer of information and matter through space.
In calculations concerning particles like protons, the speed of light is used as a conversion factor. For example, in the exercise provided, momentum was expressed in \(\text{MeV}/c\) using \(c\) as part of this relation with energy. Knowing the speed of light helps in linking mass, energy (in energy-mass equivalence), and momentum through their interactions.