Chapter 4.4, Problem 6PPB

What must the distance be between charges of +2.25 and −1.86 in order for the attractive force between them to be the same as that between charges of +4.06 and −2.11 separated by a distance of 2.16 pm?

Interpretation Introduction

Interpretation:

The distance between the two given charges using given data should be determined.

Concept Introduction:

Coulombs law: It states that the force of attraction between two point charges is directly proportional to the product of the point charges and inversely proportional the square of the distances between the charges.

$\begin{array}{l}F\propto \frac{Q_1\times Q_2}{d^{\text{2}}}\\ Q_1,Q_2\to \text{Given point charges}\\ d\to \text{distance between the charges}\end{array}$

Valence electron: The electron is considered as valence electron if it present in outermost shell of atom which gets involved in the formation of chemical bond.

Effective nuclear charge: It is the overall positive charge experienced by the outermost electrons present in the atom from the nucleus of the atom.

Screening Effect: The core electrons present near the nucleus shields the outermost electrons (valence electrons) from the charge of the nucleus.

To determine: The distance between the two given charges by using the force of attraction obtained from the two other given charges.

Answer to Problem 6PPB

Answer

The distance between two given charges is 1.51 $\text{pm}$.

Explanation of Solution

Determine the force of attraction between+4.06 and -2.11.

$\begin{array}{l}\text{The force of attraction between +4}\text{.06 and -2}\text{.11,}\\ F\propto \frac{\text{4}\text{.06}\times \left(-\text{2}\text{.11}\right)}{{\left(\text{2}\text{.16pm}\right)}^2}\\ =-1.84\end{array}$

The attraction force between charges $\text{2}\text{.25 and -1}\text{.86}$ is same as between +4.06 and -2.11 hence to obtain the distance between $\text{2}\text{.25 and -1}\text{.86}$ first force of attraction between+4.06 and -2.11 with given distance data is determined.

The attraction force is determined by substituting the given charges+4.06 and -2.11 and the distance values in the formula give the attraction force value as -1.84.

Determine the distance between the charges $\text{2}\text{.25 and -1}\text{.86}$.

$\begin{array}{l}\text{The force of attraction between +2}\text{.25 and -1}\text{.86,}\\ F\propto \frac{\text{2}\text{.25}\times \left(\text{-1}\text{.86}\right)}{{\left(\text{x}\right)}^2}\end{array}$

Given data says that attraction forces between $\text{2}\text{.25 and -1}\text{.86}$ is same as between +4.06 and -2.11. Hence, the distance between the given charges is as follows,

$\begin{array}{l}F\propto \frac{\text{2}\text{.25}\times \left(\text{-1}\text{.86}\right)}{{\left(\text{x}\right)}^2}F\propto \frac{\text{4}\text{.06}\times \left(\text{-2}\text{.11}\right)}{{\left(2.16\right)}^2}=-1.84\\ \text{It is given that,}\\ \text{Both the given charges have same attraction forces,}\\ \frac{\text{2}\text{.25}\times \left(\text{-1}\text{.86}\right)}{{\left(\text{x}\right)}^2}=-1.84\\ \frac{4.185}{{\left(\text{x}\right)}^2}=-1.84\\ {\left(\text{x}\right)}^2=2.274\\ \text{x =1}\text{.51pm}\end{array}$

The distance between the $\text{2}\text{.25 and -1}\text{.86}$ charges is obtained by equating the attraction force value with the charges substituted in given formula.

Solving for the value of x gives the distance between the two given charges.

Therefore, the distance between $\text{2}\text{.25 and -1}\text{.86}$ charges is 1.51pm.

Conclusion

Conclusion

The distance between the given charges is determined by using the coulombs formula.