Part E The ionic molecule potassium bromide (KBr), made up of a positive potassium ion (K+) of charge +e= 1.60 x 10-19 C and a negative bromine ion (Br–) of charge -e = -1.60 x 10-19 C, has an electric dipole moment of 3.50 x 10-29 Cm. Calculate the distance between the two ions. Express your answer with the appropriate units. d = 2.19×10-10 m Submit Previous Answers Part F Correct At a certain point along the axis that connects the ions, the electric field due to the KBr molecule has magnitude 8.00 x 104 N/C. How far from the center of the molecule is this point? Express your answer with the appropriate units. x = 1.99×10-8 m please show steps for how to find f and provide all the work The ionic molecule potassium bromide (KBr), made up of a positive potassium ion (K+) of charge +e=1.60×10-19C and a negative bromine ion (Br–) of charge -e=-1.60×10-19C, has an electric dipole moment of 3.50×10-29C·m. Calculate the distance between the two ions. Express your answer with the appropriate units. Part F At a certain point along the axis that connects the ions, the electric field due to the KBr molecule has magnitude 8.00×104 N/C. How far from the center of the molecule is this point? Express your answer with the appropriate units.
The correct answer and explanation is:
To find the answers, let’s break the problem down into two parts.
Part E: Finding the Distance Between the Ions
The electric dipole moment (pp) is given by: p=q⋅dp = q \cdot d
where:
- p=3.50×10−29p = 3.50 \times 10^{-29} C·m (given dipole moment)
- q=e=1.60×10−19q = e = 1.60 \times 10^{-19} C (charge of each ion)
- dd is the distance between the ions (which we need to find)
Rearranging the formula to solve for dd: d=pqd = \frac{p}{q}
Substituting the values: d=3.50×10−29 C\cdotpm1.60×10−19 Cd = \frac{3.50 \times 10^{-29} \text{ C·m}}{1.60 \times 10^{-19} \text{ C}} d=2.19×10−10 md = 2.19 \times 10^{-10} \text{ m}
So, the distance between the potassium and bromine ions is 2.19×10−102.19 \times 10^{-10} m.
Part F: Finding the Distance Where the Electric Field is Given
The electric field due to a dipole along its axis is given by: E=14πϵ0⋅2px3E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{x^3}
where:
- E=8.00×104E = 8.00 \times 10^4 N/C (given electric field)
- p=3.50×10−29p = 3.50 \times 10^{-29} C·m
- ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C²/N·m² (permittivity of free space)
- xx is the distance from the center of the dipole along its axis (which we need to find)
Rearranging the equation to solve for xx: x3=14πϵ0⋅2pEx^3 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{E}
Substituting the known values: x3=(9.0×109)×(2×3.50×10−29)8.00×104x^3 = \frac{(9.0 \times 10^9) \times (2 \times 3.50 \times 10^{-29})}{8.00 \times 10^4} x3=(9.0×109)×(7.00×10−29)8.00×104x^3 = \frac{(9.0 \times 10^9) \times (7.00 \times 10^{-29})}{8.00 \times 10^4} x3=6.30×10−198.00×104x^3 = \frac{6.30 \times 10^{-19}}{8.00 \times 10^4} x3=7.875×10−24x^3 = 7.875 \times 10^{-24}
Taking the cube root: x=(7.875×10−24)1/3x = (7.875 \times 10^{-24})^{1/3} x=1.99×10−8 mx = 1.99 \times 10^{-8} \text{ m}
So, the distance from the center of the molecule where the electric field is 8.00×1048.00 \times 10^4 N/C is 1.99×10−81.99 \times 10^{-8} m.
Explanation
The potassium bromide (KBr) molecule consists of two oppositely charged ions: a positive potassium ion (K+K^+) and a negative bromine ion (Br−Br^-). Because these charges are separated by a certain distance, the molecule forms an electric dipole. The dipole moment (pp) is a measure of the separation of charge and is calculated using the formula: p=q⋅dp = q \cdot d
where qq is the charge of each ion and dd is the separation distance. Given the dipole moment as 3.50×10−293.50 \times 10^{-29} C·m, we solve for dd, yielding 2.19×10−102.19 \times 10^{-10} m. This value represents the bond length in the KBr molecule.
In Part F, we determine the distance xx from the molecule’s center where the electric field reaches 8.00×1048.00 \times 10^4 N/C. The electric field due to a dipole along its axis is given by: E=14πϵ0⋅2px3E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{x^3}
By solving for xx, we find that the required distance is 1.99×10−81.99 \times 10^{-8} m. This shows how the dipole’s electric field weakens with distance from the molecule. Such calculations are useful in understanding molecular behavior in electric fields and in fields such as spectroscopy and electrochemistry.
Now, I will generate an image to illustrate the concept.
I’ve generated an illustration of the potassium bromide (KBr) ionic molecule, showing the charge separation, electric dipole moment, and the point along the dipole axis where the electric field is 8.00×1048.00 \times 10^4 N/C. Let me know if you need any modifications or further clarifications!
