Carbon atoms have 6 protons in their nucleus.
In nm, at what distance from a carbon nucleus is the electric potential 3.1 V?
I'm really confused with this question. I've been trying it for a while and cannot seem to get the right answer.
To determine the distance from a carbon nucleus where the electric potential is 3.1 V, we can use the formula for electric potential:
V = k * (q / r)
Where:
- V is the electric potential (in volts)
- k is the electrostatic constant (approximately 9.0 x 10^9 N*m^2/C^2)
- q is the charge (in coulombs)
- r is the distance from the charge (in meters)
In this case, the charge is the charge of the carbon nucleus, which is equivalent to the charge of a proton (q = 1.6 x 10^-19 C) because protons have a positive charge of +1.
Now, let's rearrange the formula to solve for the distance (r):
r = (k * q) / V
Substituting the given values:
r = (9.0 x 10^9 N*m^2/C^2 * 1.6 x 10^-19 C) / 3.1 V
Calculating this expression will give you the distance from the carbon nucleus at which the electric potential is 3.1 V.