Askiitians Tutor Team
Last Activity: 17 Days ago
To calculate the electric potential energy due to the electric repulsion between two carbon-12 nuclei when they are in contact, we can use the formula for electric potential energy between two point charges. The carbon-12 nucleus consists of 6 protons and 6 neutrons, so when considering the electric repulsion, we only need to focus on the protons since they carry a positive charge.
Understanding the Basics of Electric Potential Energy
The electric potential energy (U) between two point charges can be expressed with the formula:
U = k * (q1 * q2) / r
Where:
- U = electric potential energy
- k = Coulomb's constant (approximately 8.99 x 10^9 N m²/C²)
- q1 and q2 = magnitudes of the charges (in coulombs)
- r = distance between the centers of the two charges (in meters)
Calculating the Charges
For carbon-12, each nucleus has 6 protons. The charge of a single proton is approximately +1.6 x 10^-19 C. Therefore, the total charge of one carbon-12 nucleus is:
q1 = q2 = 6 protons * 1.6 x 10^-19 C/proton = 9.6 x 10^-19 C
Determining the Distance
When two nuclei touch, we can estimate the distance between their centers. The radius of a typical atomic nucleus is about 3 x 10^-15 m (or 3 femtometers). Since we are considering two nuclei, the distance between their centers when they touch is approximately:
r = 2 * radius of one nucleus = 2 * 3 x 10^-15 m = 6 x 10^-15 m
Plugging in the Values
Now we can substitute the values into the electric potential energy formula:
U = (8.99 x 10^9 N m²/C²) * (9.6 x 10^-19 C * 9.6 x 10^-19 C) / (6 x 10^-15 m)
Calculating the Result
First, calculate the product of the charges:
q1 * q2 = (9.6 x 10^-19 C) * (9.6 x 10^-19 C) = 9.216 x 10^-37 C²
Now, substituting this back into the equation:
U = (8.99 x 10^9 N m²/C²) * (9.216 x 10^-37 C²) / (6 x 10^-15 m)
Calculating this gives:
U ≈ (8.99 x 10^9 * 9.216 x 10^-37) / (6 x 10^-15)
U ≈ 1.37 x 10^-13 J
Final Thoughts
The electric potential energy due to the electric repulsion between two carbon-12 nuclei when they touch is approximately 1.37 x 10^-13 joules. This energy is significant in nuclear physics, especially when considering nuclear reactions and the forces at play within atomic nuclei.