Question

The nucleus of a carbon- 14 atom (mass \(=14\) amu) has diameter of \(3.01 \mathrm{fm} .\) It has 6 protons and a charge of \(+6 e .\) a) What is the force on a proton located at \(3 \mathrm{fm}\) from the surface of this nucleus? Assume that the nucleus is a point charge. b) What is the proton's acceleration?

Short answer

Answer: The force on the proton is 3.78 x 10^-10 N, and its acceleration is 2.26 x 10^17 m/s^2.

Step-by-step solution

a) Finding the Force on a Proton

We are given the charge of the carbon-14 nucleus, which is \(q_{1}=+6e\), and the charge of a proton, which is \(q_{2}=+ e\). We are also given the diameter of the carbon-14 nucleus \(d = 3.01\,\text{fm}\) and the distance from the proton to the nucleus' surface \(r_{\text{surface}}=3\,\text{fm}\). The distance between their centers is the sum of these distances: $$r = \frac{d}{2} + r_{\text{surface}}$$ Now, we can find the distance \(r\) in meters: $$r = \frac{3.01\,\text{fm}}{2} + 3\,\text{fm} = 4.505\,\text{fm} = 4.505 \times 10^{-15}\,\text{m}$$ Using Coulomb's law formula, we find the force on the proton: $$F = k \frac{ q_{1} q_{2} }{ r^{2} } = \frac{ (8.9875 \times 10^{9} \,\text{N m^{2} C^{-2}}) (6e)(e) }{ (4.505 \times 10^{-15} \,\text{m})^{2} }$$ Calculating the force, we get: $$F = 3.78 \times 10^{-10}\, \mathrm{N}$$

b) Finding the Proton's Acceleration

Now we will find the acceleration of the proton using Newton's second law of motion. We are given the mass of a proton \(m_{p}= 1.673\times10^{-27}\,\text{kg}\). The formula for acceleration is: $$a = \frac{F}{m_{p}}$$ Plugging in the values for the force and mass of the proton, we get: $$a = \frac{3.78 \times 10^{-10} \, \mathrm{N}}{1.673 \times 10^{-27} \, \mathrm{kg}}$$ Calculating the acceleration, we get: $$a = 2.26\times10^{17}\,\mathrm{m/s^{2}}$$ The force on the proton located at \(3\,\text{fm}\) from the surface of the carbon-14 nucleus is \(3.78\times10^{-10}\,\text{N}\), and its acceleration is \(2.26\times10^{17}\,\text{m/s^{2}}\).

Key concepts

Nuclear Physics

Nuclear physics is a branch of physics that deals with the constituents and interactions of atomic nuclei. It is essential for understanding the properties of matter at the smallest scale and the forces that bind protons and neutrons in the nucleus.

Protons, just like the ones in a carbon-14 nucleus, are part of this subatomic world. They carry a positive electric charge and are held tightly in the nucleus by the nuclear force, which is distinct from the electromagnetic force that governs the interaction of electric charges in Coulomb's Law. Knowledge of nuclear physics allows us to calculate the behavior of these protons when subjected to various forces, whether within a nucleus or when interacting with external charged particles.

Force Calculation

Force calculation is a fundamental aspect of physics that involves determining the amount of push or pull on an object. In the context of charged particles, Coulomb's Law provides a precise formula to calculate the electric force between two point charges:
\requation: \( F = k \frac{ q_1 q_2 }{ r^2 } \)
where \(F\) is the electric force, \(k\) is Coulomb's constant (approximately \(8.9875 \times 10^9 \mathrm{Nm^2C^{-2}}\)), \(q_1\) and \(q_2\) are the electric charges, and \(r\) is the distance between the charges. This allows us to understand the strength of electromagnetic interaction that could cause proton acceleration, for instance.

Electric Charge

Electric charge is a fundamental property of particles that causes them to experience a force when placed in an electric and magnetic field. There are two kinds of electric charges: positive and negative. Protons have a positive charge denoted by \(+e\), where \(e\) represents the elementary charge equal to approximately \(1.602 \times 10^{-19}\) coulombs.

In a nucleus like that of carbon-14, the protons create a net positive charge that can influence other charges around it. Understanding electric charge is key to predicting the behavior of charged particles, such as the force between two protons using Coulomb's Law.

Proton Acceleration

Proton acceleration is the process by which a proton gains velocity due to an applied force. In the case of a proton near a carbon-14 nucleus, the force calculated through Coulomb's Law can cause significant acceleration. Newton's second law of motion, given by \( a = \frac{F}{m} \), tells us that the acceleration \(a\) of an object is directly proportional to the net force \(F\) acting on it and inversely proportional to its mass \(m\).

Even though a proton is incredibly small, with a mass of approximately \(1.673 \times 10^{-27}\) kilograms, the forces at the nuclear level are so tremendous that they can result in startlingly high acceleration values, which are much larger than we experience in our everyday macroscopic world.