To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter ¹²C nucleus. The proton must impact the nucleus with a kinetic energy of 3.00 MeV. Assume the nucleus remains at rest. Through what potential difference must the proton be accelerated from rest to acquire this speed?

Verified step by step guidance

1. Determine the relationship between the kinetic energy of the proton and the potential difference it must be accelerated through. The work-energy principle states that the work done on the proton by the electric field is equal to its kinetic energy. This can be expressed as: $K\; =\; q\; \backslash (\backslash cdot\backslash )\; \backslash (\backslash Delta\backslash )\; V$, where $K$ is the kinetic energy, $q$ is the charge of the proton, and $\backslash (\backslash Delta\backslash )\; V$ is the potential difference.
2. Rearrange the formula to solve for the potential difference: $\backslash (\backslash Delta\backslash )\; V\; =\; \backslash (\backslash frac\{K\}\{q\}\backslash )$. Here, $K\; =\; 3.00\; \backslash \; \backslash (\backslash text\{MeV\}\backslash )$ (convert this to joules for consistency in SI units), and $q$ is the charge of the proton, which is $1.602\; \backslash (\backslash times\backslash )\; 10^\{-19\}\; \backslash \; \backslash (\backslash text\{C\}\backslash )$.
3. Convert the kinetic energy from MeV to joules. Use the conversion factor $1\; \backslash \; \backslash (\backslash text\{MeV\}\backslash )\; =\; 1.602\; \backslash (\backslash times\backslash )\; 10^\{-13\}\; \backslash \; \backslash (\backslash text\{J\}\backslash )$. Multiply $3.00\; \backslash \; \backslash (\backslash text\{MeV\}\backslash )$ by this factor to get the kinetic energy in joules.
4. Substitute the values of $K$ (in joules) and $q$ (in coulombs) into the formula $\backslash (\backslash Delta\backslash )\; V\; =\; \backslash (\backslash frac\{K\}\{q\}\backslash )$ to calculate the potential difference.
5. Ensure the units are consistent and verify the calculation. The resulting potential difference will be in volts, which represents the energy per unit charge required to accelerate the proton to the given kinetic energy.

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 1/2 mv², where m is mass and v is velocity. In nuclear physics, the kinetic energy of particles like protons is crucial for understanding their ability to overcome potential barriers, such as the Coulomb barrier in nuclear reactions.

Potential Difference

Potential difference, or voltage, is the work done per unit charge to move a charge between two points in an electric field. It is directly related to the kinetic energy gained by a charged particle when accelerated through an electric field, expressed as KE = qV, where q is the charge and V is the potential difference.

Coulomb Barrier

The Coulomb barrier is the energy barrier due to electrostatic repulsion that charged particles must overcome to get close enough for nuclear reactions to occur. In this context, the proton must have sufficient kinetic energy to overcome this barrier to successfully impact the nucleus of the carbon atom.