Answer :
The potassium atom is approximately [tex]\rm 1.79 \times 10^{-10}[/tex] meters from the center of mass in the KI molecule.
To find the distance of the potassium atom from the center of mass in the potassium iodide (KI) molecule, we can use the concept of reduced mass (μ) and the formula for the centre of mass position.
The reduced mass of the system is given by:
[tex]\rm \[\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}\][/tex]
Where:
[tex]\rm \(m_1\)[/tex] is the mass of the potassium atom (39.1 amu converted to kg),
[tex]\rm \(m_2\)[/tex] is the mass of the iodine atom (127 amu converted to kg).
Now, calculate μ:
[tex]\[\mu = \frac{(39.1 \times 1.67 \times 10^{-27} kg) \cdot (127 \times 1.67 \times 10^{-27} kg)}{(39.1 \times 1.67 \times 10^{-27} kg) + (127 \times 1.67 \times 10^{-27} kg)}\\\mu = \(1.45 \times 10^{-27} kg\)[/tex]
Now, we can use the formula for the center of mass position:
[tex]\[R_{\text{center of mass}} = \frac{m_1 \cdot r_1 + m_2 \cdot r_2}{m_1 + m_2}[/tex]
Where:
[tex]\rm \(m_1\)[/tex] and [tex]\rm \(m_2\)[/tex] are the masses of the atoms,
[tex]\rm \(r_1\)[/tex] and [tex]\rm \(r_2\)[/tex] are their respective distances from the center of mass.
We're solving for [tex]\rm \(r_1\)[/tex] (the distance of the potassium atom from the center of mass), so rearrange the formula:
[tex]\rm \[r_1 = \frac{R_{\text{center of mass}} \cdot (m_1 + m_2) - m_2 \cdot r_2}{m_1}[/tex]
Now, substitute the known values:
[tex]\[r_1 = \frac{0 \cdot (39.1 \times 1.67 \times 10^{-27} kg) - (127 \times 1.67 \times 10^{-27} kg) \cdot 2.79 \times 10^{-10} m}{39.1 \times 1.67 \times 10^{-27} kg}\][/tex]
Calculate [tex]\rm \(r_1\)[/tex]:
[tex]\rm \[r_1 = -1.79 \times 10^{-10} m\][/tex]
So, the potassium atom is approximately [tex]\rm 1.79 \times 10^{-10}[/tex] meters from the center of mass in the KI molecule.
Know more about potassium atom:
https://brainly.com/question/33440930
#SPJ1
