Question

The nucleus of a helium atom contains two protons that are approximately 1 fm apart. How much work must be done by an external agent to bring the two protons from an infinite separation to a separation of \(1.0 \mathrm{fm} ?\)

Short answer

Answer: The work done to bring the protons is approximately \(2.31 * 10^{-28} J\).

Step-by-step solution

Understand the Coulomb's law potential energy formula

Coulomb's law can be used to calculate the potential energy between two charged particles. The potential energy (U) is given by the formula: \[U = k * \frac{q_1 * q_2}{r}\] where k is the Coulomb's constant (approximately equal to \(8.9875517923*10^9 N m^2 C^{-2}\)), \(q_1\) and \(q_2\) are the charges of the particles, and r is the distance between the two particles.

Determine the charges and distance at infinite separation

First, we need to find the charges of the two protons and the initial distance between them. A proton has a charge of approximately \(+1.602176634*10^{-19} C\). When the protons are at an infinite separation, their potential energy is essentially 0, because the distance between them is so large that it has a negligible effect on their potential energy.

Calculate the potential energy at 1.0 fm separation

Next, we need to calculate the potential energy of the protons when they are at a separation of 1.0 fm (1 femtometer = \(10^{-15} m\)). Using the same Coulomb's law potential energy formula from Step 1, replace r with \(1.0*10^{-15} m\) and the charges \(q_1\) and \(q_2\) with the charge of a proton we obtained in Step 2. Plugging these values into the formula, we get: \[U = 8.9875517923*10^9 * \frac{(1.602176634*10^{-19})^2}{1.0*10^{-15}}\]

Calculate the work done by the external agent

The work done by an external agent is the difference between the potential energies at the initial and final separations. Since the potential energy is 0 at infinite separation, the work done is equal to the potential energy at 1.0 fm separation: \[W = U_{1.0fm} - U_{\infty} = U_{1.0fm}\] Now, calculate the value we obtained in Step 3: \[W = 8.9875517923*10^9 * \frac{(1.602176634*10^{-19})^2}{1.0*10^{-15}}\]

Solve for the work done by the external agent

Plug the given values into a calculator to solve for the work done: \[W = 2.3069 * 10^{-28} J\]

State the answer

The work that must be done by an external agent to bring the two protons from an infinite separation to a separation of 1.0 fm is approximately \(2.31 * 10^{-28} J\).