Question

(a) What is a mass of \(1 \mathrm{MeV} / c^{2}\) in kilograms? ( \(\mathbf{b}\) ) What is a momentum of \(1 \mathrm{MeV} / c\) in \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s} ?\)

Short answer

a) Mass is approximately \(1.783 \times 10^{-30}\, \text{kg}.\) b) Momentum is approximately \(5.34 \times 10^{-22}\, \text{kg} \cdot \text{m/s}.\)

Step-by-step solution

Understand the units and constants

The exercise involves converting energy units to mass and momentum measurements. **MeV** (Mega electron volts) is a unit of energy. You'll need to use the speed of light, denoted as **c**, where \( c = 3.00 \times 10^8 \text{ m/s} \). You also need the conversion factor between energy in MeV and mass in kg, which involves the equivalence \( 1 \text{eV} = 1.60219 \times 10^{-19} \text{ J} \), and the relation \( E = mc^2 \).

Convert MeV to Joules

To convert from MeV to Joules (J), use: \( 1 \text{MeV} = 1.60219 \times 10^{-13} \text{ J} \) (since 1 MeV = 1 million eV).

Use mass-energy equivalence

Using the relation \( E = mc^2 \), rearrange to find mass \( m \) as \( m = \frac{E}{c^2} \). Substitute \( E = 1.60219 \times 10^{-13} \text{ J} \) and \( c = 3.00 \times 10^8 \text{ m/s} \).

Calculate mass

Now compute: \[ m = \frac{1.60219 \times 10^{-13}}{(3.00 \times 10^8)^2} \approx 1.783 \times 10^{-30} \text{ kg} \]

Convert MeV/c to kg·m/s for momentum

For momentum, \( 1 \text{ MeV}/c \) describes the momentum value in energy units over speed of light. Use the conversion to Joules: \( 1.60219 \times 10^{-13} \text{ J} \).

Calculate momentum

Calculate momentum as: \[ p = \frac{1.60219 \times 10^{-13}}{3.00 \times 10^8} \approx 5.34 \times 10^{-22} \text{ kg} \cdot \text{m/s} \]

Key concepts

Unit Conversion

Converting units is crucial when dealing with physics problems. Here, we want to change units from Mega electron volts (MeV) to kilograms (kg) and kilogram meters per second (kg·m/s). MeV is a measure of energy that's often used in particle physics. Each MeV is equivalent to \(1.60219 \times 10^{-13}\) Joules (J).
To solve the mass-energy equivalence problem, we first need to convert from MeV to J. Knowing the energy in Joules allows us to use the equation \(E = mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light.
Remember:

• \(1 \text{ MeV} = 1.60219 \times 10^{-13} \text{ J}\)
• \(c = 3.00 \times 10^8 \text{ m/s}\)

This knowledge helps in converting between energy units and other physical quantities.

Momentum Calculation

Momentum is a key concept in physics, representing the quantity of motion an object has. When energy is expressed in MeV and divided by the speed of light \(c\), it gives a momentum value. This is particularly useful in particle physics.
For our calculations, \(1 \text{ MeV}/c\) indicates momentum in energy units over speed, simplifying the direct use of the conversion factor.
The equation to calculate momentum is: \( p = \frac{E}{c} \), where:

• \(p\) is the momentum
• \(E\) is the energy in Joules
• \(c\) is the speed of light

By substituting the known values, \(E = 1.60219 \times 10^{-13} \text{ J}\) and \(c = 3.00 \times 10^8 \text{ m/s}\), we find the momentum to be approximately \(5.34 \times 10^{-22} \text{ kg} \cdot \text{m/s}\).
This calculation gives us the momentum in standard SI units.

Speed of Light

The speed of light, denoted as \(c\), is a fundamental constant in physics and key to understanding mass-energy equivalence. It is approximately \(3.00 \times 10^8 \text{ m/s}\). In the famous equation \(E = mc^2\), \(c\) represents the speed at which light travels in a vacuum.
This speed is crucial when converting energy units into mass, or calculating momentum, because it bridges the gap between what we know about energy, mass, and motion.

• In mass-energy equivalence, the speed of light squared \(c^2\) converts energy in Joules to mass in kilograms.
• In momentum calculations, \(c\) helps determine the momentum of an object based on its energy content.

Understanding the role of the speed of light in these calculations helps clarify how fundamental physics concepts intertwine.