Question

A sodium ion (m=22.99g/mol, q=+1.6x10-19C) in the crystal structure of table sale lies .236nm from a chloride ion (35.45g/mol, q= -1.6x10-19C). Find the mutual force between the ions.

Answer 1

Answer on Question 58951, Physics, Electromagnetism

Question:

A sodium ion $(m = 22.99\ g/mol, q = 1.6 \cdot 10^{-19}\ C)$ in the crystal structure of table solt lies $0.236\ nm$ from a chloride ion $(m = 35.45\ g/mol, q = -1.6 \cdot 10^{-19}\ C)$. Find the mutual force between the ions.

Solution:

We can find the electric force (mutual force) between the ions from the Coulomb's law:

$$F_e = k \frac{|q_1 q_2|}{r^2},$$

here, $q_1 = 1.6 \cdot 10^{-19} \, \text{C}$ is the charge of the sodium ion, $q_2 = -1.6 \cdot 10^{-19} \, \text{C}$ is the charge of the chloride ion, $r$ is the distance between two ions, $k$ is the Coulomb's constant.

Let's substitute the numbers:

$$F_e = k \frac{|q_1 q_2|}{r^2} = 9 \cdot 10^9 \frac{N \cdot m^2}{C^2} \cdot \frac{|(1.6 \cdot 10^{-19} \, \text{C}) \cdot (-1.6 \cdot 10^{-19} \, \text{C})|}{(0.236 \cdot 10^{-9} \, m)^2} = 4.14 \cdot 10^{-9} \, N.$$

Answer:

$$F_e = 4.14 \cdot 10^{-9} \, N.$$

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