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?

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: $$F = 3.78 \times 10^{-10}\, \mathrm{N}$$

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 \(2.26\times10^{17}\,\text{m/s^{2}}\).