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 \(3.00 \mathrm{fm}\) 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 approximately \(2.14 \times 10^{-9} N\), and its acceleration is approximately \(1.28 \times 10^{18} m/s^2\).

Step-by-step solution

Calculate the electric force

To find the electric force on the proton, we will use Coulomb's law, which states that the electrostatic force (F) acting between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula for Coulomb's law is: $$ F = k \frac{q_1q_2}{r^2} $$ Here, \(k=8.99\times 10^{9} Nm^2 C^{-2}\) is the electrostatic constant, \(q_1\) and \(q_2\) are the two charges, and \(r\) is the distance between the charges. In our case, the charge of the nucleus, \(q_1 = +6e\), and the charge of the proton, \(q_2 = +e\). The distance between them, \(r\), is the diameter of the nucleus (3.01 fm) plus the distance from the surface (3.00 fm), which gives us a total distance of 6.01 fm. However, we need to convert the distance from femtometers (fm) to meters (m) for the calculation, since the constant k is in Nm^2 C^{-2}. To do this, we use the conversion factor 1 fm = \(10^{-15} m\). $$ r = (6.01 \mathrm{fm}) (\frac{10^{-15} m}{1 \mathrm{fm}}) = 6.01\times 10^{-15} m$$ Now, we can plug the values into the formula to find the electric force: $$ F = (8.99\times 10^{9} Nm^2 C^{-2}) \frac{(+6e)(+e)}{(6.01\times 10^{-15} m)^2} $$ $$ F = 2.14 \times 10^{-9} N $$

Calculate the proton's acceleration

Now that we have the electric force on the proton, we can use Newton's second law of motion to find the proton's acceleration (a). The formula is: $$ F = m_p \times a $$ We know the force (F) on the proton, and the mass of a proton (\(m_p\)) is approximately \(1.67\times 10^{-27} kg\). We can solve for the acceleration (a) by dividing both sides by the mass of the proton. $$ a = \frac{F}{m_p} = \frac{2.14 \times 10^{-9} N}{1.67\times 10^{-27} kg}$$ $$ a = 1.28 \times 10^{18} m/s^2 $$ So the acceleration of the proton is approximately \(1.28 \times 10^{18} m/s^2\). In summary, the force on the proton is \(2.14 \times 10^{-9} N\), and its acceleration is \(1.28 \times 10^{18} m/s^2\).