Question

are charges attractive or repulsive?
the force between two particles of opposite charges — one charge is positive and the other is negative — is attraction.
the force between two particles of the same charge — both particles are positive, or both particles are negative — is repulsion.
why is coulomb’s law useful?
coulomb’s law is often used to describe the force between protons and electrons in a single atom. this can help scientists predict how electrons will behave and whether or not they will stay in the atom.
let’s use coulomb’s law to characterize the forces between a sodium atom’s nucleus and an electron in its valence shell.
a sodium atom has 11 protons in its nucleus, each with a charge of +1 e. the nucleus has a charge of +11 elementary charges, so ( q_1 = 11 , e ).
each electron in the valence shell has a charge of -1, so ( q_2 = -1 , e ).
the nucleus is 227 pm from the sodium’s outermost electron shell (figure 1), so ( r = 227 , \text{pm} ).
figure 1: a diagram of a sodium atom. the distance between sodium’s outermost electron shell and its nucleus is 227 pm.
coulomb’s law for the force between a sodium atom’s nucleus and an electron in its outer shell is below:
f = \frac{(0.0000000 \frac{n cdot \text{pm}^2}{e^2}) , (+11e) , (-1e)}{(227 , \text{pm})^2}

Explanation:

Step1: Recall Coulomb's Law Formula

Coulomb's law is given by \( F = k\frac{q_1q_2}{r^2} \), where \( k = 0.0000000000000000000000000000000008988\frac{N\cdot m^2}{C^2} \) (or in terms of elementary charges and picometers, we can use the given form \( F=\frac{(0.0000000000000000000000000000000008988\frac{N\cdot pm^2}{e^2})q_1q_2}{r^2} \) since \( 1\ C = 6.2415\times10^{18}\ e \) and \( 1\ m = 10^{12}\ pm \), so converting units gives \( k\) in \( \frac{N\cdot pm^2}{e^2} \) as approximately \( 0.0000000000000000000000000000000008988\frac{N\cdot pm^2}{e^2} \)). Here, \( q_1 = 11e \), \( q_2=- 1e \), and \( r = 227\ pm \).

Step2: Substitute Values into the Formula

Substitute \( q_1 = 11e \), \( q_2=-1e \), and \( r = 227\ pm \) into the formula:

$$ F=\frac{(0.0000000000000000000000000000000008988\frac{N\cdot pm^2}{e^2})(11e)(-1e)}{(227\ pm)^2} $$

First, calculate the numerator: \( (0.0000000000000000000000000000000008988)(11)(- 1)e^2=- 0.0000000000000000000000000000000098868e^2\frac{N\cdot pm^2}{e^2}=- 0.0000000000000000000000000000000098868N\cdot pm^2 \)
Denominator: \( (227)^2\ pm^2 = 51529\ pm^2 \)

Step3: Calculate the Force

Now, divide the numerator by the denominator:

$$ F=\frac{- 0.0000000000000000000000000000000098868N\cdot pm^2}{51529\ pm^2} $$

$$ F\approx - 1.92\times10^{-36}\ N $$

The negative sign indicates an attractive force (since opposite charges attract), which makes sense as the nucleus (positive) and electron (negative) should attract.

Answer:

The force between the sodium atom's nucleus and the valence electron is approximately \( \boldsymbol{-1.92\times10^{-36}\ N} \) (the negative sign indicates attraction).