Chapter 5: Problem 60
Find the rest energy in joules and MeV of a proton, given its mass is \(1.67 \times 10^{-27} \mathrm{kg}\).
Short Answer
Expert verified
The rest energy of a proton is approximately \(1.503 \times 10^{-10}\) Joules or 939.3 MeV.
Step by step solution
01
Recall the formula for rest energy
The formula for rest energy is given by Einstein's equation:
\[E=mc^2\]
Where E is the rest energy, m is the mass (in kg), and c is the speed of light (\( c=3 \times 10^8 \mathrm{m/s}\)).
02
Calculate the rest energy in Joules
Now we will calculate the rest energy (E) of a proton in Joules using the given mass and the speed of light.
\[m = 1.67 \times 10^{-27} \mathrm{kg}\]
\[c = 3 \times 10^8 \mathrm{m/s}\]
\[E = mc^2\]
\[E = (1.67 \times 10^{-27} \mathrm{kg}) (3 \times 10^8 \mathrm{m/s})^2\]
Now calculate the result:
\[E ≈ 1.503 \times 10^{-10} \mathrm{J}\]
So the rest energy of a proton is approximately \(1.503 \times 10^{-10}\) Joules.
03
Convert rest energy to MeV
To convert the rest energy from Joules to MeV (Mega electron-volts), we can use the following conversion factor:
\[1 \mathrm{J} = 6.242 \times 10^{12} \mathrm{MeV}\]
Using this conversion factor, we can find the rest energy of a proton in MeV:
\[E \approx 1.503 \times 10^{-10} \mathrm{J} \times \frac{6.242 \times 10^{12} \mathrm{MeV}}{1 \mathrm{J}}\]
Now, calculate the result:
\[E ≈ 939.3 \mathrm{MeV}\]
So, the rest energy of a proton is approximately 939.3 MeV.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Einstein's equation E=mc^2
Albert Einstein's iconic equation, E=mc^2, is at the heart of modern physics, encapsulating a profound principle that links matter and energy. The 'E' in this equation represents energy, 'm' stands for mass, and 'c' denotes the speed of light in a vacuum, which is roughly 3 x 10^8 meters per second.
This equation tells us that mass can be converted into energy and vice versa, which means a small amount of mass can potentially release a vast amount of energy. This is the fundamental principle behind nuclear power and atomic bombs.
It's crucial to note that the energy obtained from mass as per this equation is known as 'rest energy', which is the energy possessed by an object at rest. The remarkable aspect of this equation is that it works universally, whether you're looking at an apple, a pencil, or a subatomic particle like a proton.
In layman's terms, the equation implies that an object with mass has an intrinsic energy – its rest energy – even when it's not in motion.
This equation tells us that mass can be converted into energy and vice versa, which means a small amount of mass can potentially release a vast amount of energy. This is the fundamental principle behind nuclear power and atomic bombs.
It's crucial to note that the energy obtained from mass as per this equation is known as 'rest energy', which is the energy possessed by an object at rest. The remarkable aspect of this equation is that it works universally, whether you're looking at an apple, a pencil, or a subatomic particle like a proton.
In layman's terms, the equation implies that an object with mass has an intrinsic energy – its rest energy – even when it's not in motion.
Rest energy calculation
Calculating the rest energy of an object, according to Einstein's equation, requires knowledge of the object’s mass and the speed of light. The equation, as we've stated, is E=mc^2, where 'E' is the rest energy, 'm' is the mass, and 'c' is the speed of light. To calculate the rest energy of a proton, we insert its mass into this equation alongside the squared value of light speed.
A proton's mass is approximately 1.67 x 10^-27 kilograms. Squaring the constant speed of light (3 x 10^8 meters per second) and multiplying it by the proton's mass gives us the rest energy. The result, through the calculation steps provided, reveals the rest energy to be approximately 1.503 x 10^-10 Joules.
This calculation is paramount in understanding just how much energy is tied up in even small amounts of mass – exemplifying the power of Einstein's theory.
A proton's mass is approximately 1.67 x 10^-27 kilograms. Squaring the constant speed of light (3 x 10^8 meters per second) and multiplying it by the proton's mass gives us the rest energy. The result, through the calculation steps provided, reveals the rest energy to be approximately 1.503 x 10^-10 Joules.
This calculation is paramount in understanding just how much energy is tied up in even small amounts of mass – exemplifying the power of Einstein's theory.
Energy conversion joules to MeV
To relate the rest energy of particles like protons to quantities used in particle physics, we convert the units from joules to mega electron-volts (MeV). The conversion factor between these units is that 1 joule equals approximately 6.242 x 10^12 MeV.
Using this conversion factor, we can translate the rest energy we computed in joules to a more convenient unit in particle physics contexts. For the given proton mass, the rest energy in joules is approximately 1.503 x 10^-10 J. Multiplying this number by the conversion factor yields the energy in MeV.
The conversion demonstrates that the rest energy of a proton is around 939.3 MeV. This value is widely recognized by physicists, as it highlights the mass-energy relationship at a scale where quantum effects and particle interactions are significant. By understanding and using these conversions, one can better appreciate the energy scales involved in subatomic phenomena.
Using this conversion factor, we can translate the rest energy we computed in joules to a more convenient unit in particle physics contexts. For the given proton mass, the rest energy in joules is approximately 1.503 x 10^-10 J. Multiplying this number by the conversion factor yields the energy in MeV.
The conversion demonstrates that the rest energy of a proton is around 939.3 MeV. This value is widely recognized by physicists, as it highlights the mass-energy relationship at a scale where quantum effects and particle interactions are significant. By understanding and using these conversions, one can better appreciate the energy scales involved in subatomic phenomena.