Answer
64.8k+ views
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.
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)