Question

In a CO molecule, the distance between C (mass=12 amu) and O (mass=16 amu) is;
Where 1 amu= $\dfrac{5}{3} \times {10^{ - 27}}kg$
(A) $2.4 \times {10^{ - 10}}m$
(B) $1.9 \times {10^{ - 10}}m$
(C) $1.3 \times {10^{ - 10}}m$
(D) $4.4 \times {10^{ - 11}}m$

Answer 1

Hint: the moment of inertia of the molecule CO is${I_{CO}} = 1.8 \times {10^{ - 47}}kg{m^2}$. The moment of inertia about the axis passing through the center of mass will help us get the distance between the two atoms.

Complete solution:
The CO molecule consists of two atoms of different masses. They are C (carbon) and O (oxygen). This tells that the center of mass is not in the middle but somewhere other than the midpoint.
Center of mass of a system of two or masses is the point at which all mass is concentrated, this assumption can be made. It is given by;
Here, $\bar x$ is the x coordinate of the center of mass. ${m_1},{m_2},....$ are masses of individual elements and ${x_1},{x_2},....$ are their positions with respect to center of mass respectively.
Hence in this case we first find the distance of both the atoms from the center of mass, the distances can be given by;
${x_C} = \dfrac{{{m_O}d}}{{{m_C} + {m_O}}},{x_O} = \dfrac{{{m_C}d}}{{{m_C} + {m_O}}}$ here d is the total distance between the atoms.
So, ${x_C} = \dfrac{{16d}}{{12 + 16}} = \dfrac{{16d}}{{28}}$ (equation: 1)
Also, ${x_o} = \dfrac{{12d}}{{12 + 16}} = \dfrac{{12d}}{{28}}$ (equation: 2)
Now if we talk about the moment of inertia about the axis passing through the center of mass, we know that by definition moment of inertia about the axis pass through center will be given by;
${I_{CO}} = {m_C}{r_C} + {m_O}{r_O}$
Substituting the value of equation: 1 and equation: 2 in the above equation we get
${I_{CO}} = 12\left( {\dfrac{{16d}}{{28}}} \right) + 16\left( {\dfrac{{12d}}{{28}}} \right)$
${I_{CO}} = \dfrac{{96d}}{7}$
Now this is amu-square meter, we need to convert it to kg-square meter.
We are given that, $1amu = \dfrac{5}{3} \times {10^{ - 27}}kg$.
So, ${I_{CO}} = \dfrac{{96d}}{7} \times \dfrac{5}{3} \times {10^{ - 27}}kg{m^2}$.
Also we know it as a fact that ${I_{CO}} = 1.8 \times {10^{ - 47}}$
Hence, ${I_{CO}} = \dfrac{{96d}}{7} \times \dfrac{5}{3} \times {10^{ - 27}} = 1.8 \times {10^{ - 47}}$
Solving the above equation for d we get;
$d = 1.3 \times {10^{ - 10}}m$

Hence option C is correct.

Note:(1) We assume that these distances are measured from the center of the atom. So, these measurements consist of the atomic radii as well.
(2) It is logical to consider the measurements from the atomic center point because the atom's mass is concentrated at that point.