Calculate the electrical attraction that a proton in a nucleus exerts on an orbiting electron if the two particles are 1.13×10^-10 m apart.
Use Coulomb's Law:
F = k*Q1*Q2 / r^2
k = 9.00x10^9 N.m^2/C^2 (Coulomb's constant)
Q1 = Q2 = 1.60x10^-19C (Elementary charge)
r = given in the question (distance)
Note: The charge of a proton is equal to the charge of an electron except for the algebraic sign.
I tried that and got 1.8x10^8 but it wasn't the right answer
You didn't do it right. Just doing the orders of magnitude only..
F=E9*E-19*E-19/E-20 ( I squared the distance)
F=E(9-38+20) appx 10^-9 Newtons.
what does E mean? Could you explain it another way because I'm confused.
To calculate the electrical attraction between a proton and an electron, you can use Coulomb's Law. Coulomb's Law states that the electrical attraction between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = (k * q₁ * q₂) / r²
Where:
F is the force of attraction between the charges
k is the electrostatic constant (9.0 x 10^9 N m²/C²)
q₁ is the charge of the first particle (proton in this case)
q₂ is the charge of the second particle (electron in this case)
r is the distance between the particles
In our case, the proton has a charge of +1.6 x 10^-19 C (Coulombs) and the electron has a charge of -1.6 x 10^-19 C. The distance between them is 1.13 x 10^-10 m.
Plugging these values into the formula, we have:
F = (9.0 x 10^9 N m²/C² * (1.6 x 10^-19 C)^2) / (1.13 x 10^-10 m)²
Simplifying the equation using the given values:
F = (9.0 x 10^9 N m²/C² * 2.56 x 10^-38 C²) / 1.2769 x 10^-20 m²
F = (2.304 x 10^-28 N m²) / 1.2769 x 10^-20 m²
F ≈ 1.803 x 10^-8 N
Therefore, the electrical attraction between the proton and the orbiting electron is approximately 1.803 x 10^-8 Newtons.