Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 553

In HCl molecule the separation between the nuclei of the two atoms is about \(1.27 \mathrm{~A}\left(1 \mathrm{~A}=10^{-10}\right)\). The approximate location of the centre of mass of the molecule is \(-\mathrm{A}\) i \(\wedge\) with respect of Hydrogen atom (mass of CL is \(35.5\) times of mass of Hydrogen \()\) \(\\{\mathrm{A}\\} 1 \mathrm{i}\) \\{B \\} \(2.5 \mathrm{i}\) \\{C \(\\} 1.24 \mathrm{i}\) \\{D \(1.5 \mathrm{i}\)

Short Answer

Expert verified
The position of the center of mass of the HCl molecule with respect to the hydrogen atom is approximately \(1.24 Å\), which corresponds to option C.

Step by step solution

01

Understanding the center of mass formula

The center of mass (CM) of a two-particle system can be found using the following formula: CM \(=\frac{m_1x_1 + m_2x_2}{m_1 + m_2}\) where: - m1 and m2 are the masses of the two particles, - x1 and x2 are the positions of the two particles, and - CM is the position of the center of mass. In this case, we have only one dimension (the x-axis), so the formula can be simplified to: CM \(= \frac{m_Hx_H + m_{Cl}x_{Cl}}{m_H + m_{Cl}}\) where: - \(m_H\) is the mass of the hydrogen atom, - \(m_{Cl}\) is the mass of the chlorine atom, - \(x_H\) is the position of the hydrogen atom, and - \(x_{Cl}\) is the position of the chlorine atom.
02

Assigning values to the variables

Since we are calculating the position of the center of mass with respect to the hydrogen atom, we can set the position of hydrogen atom as the origin. Thus, \(x_H = 0\). Now, we only need to find the position of the chlorine atom (\(x_{Cl}\)) and the ratio of the masses of chlorine and hydrogen atoms (\(m_H\) and \(m_{Cl}\)) Given, the separation between the nuclei of the two atoms is: \(x_{Cl} = 1.27 Å\) Since \(1 Å = 10^{-10}\), the separation between the nuclei of the two atoms is: \(x_{Cl} = 1.27 * 10^{-10} meters\) Also, given that the mass of the chlorine atom is 35.5 times the mass of the hydrogen atom: \(m_{Cl} = 35.5 * m_H\)
03

Calculating the position of the center of mass

Now, we can use the formula for the center of mass to find the position of the center of mass: CM \(= \frac{m_H*0 + (35.5*m_H)(1.27*10^{-10})}{m_H + 35.5*m_H}\) As we can observe, the mass of the hydrogen atom (\(m_H\)) cancels out in the numerator and denominator: CM \(= \frac{(35.5)(1.27*10^{-10})}{1 + 35.5}\) On calculating the position of the center of mass, we get: CM \(= 1.2444 * 10^{-10} meters\)
04

Converting the center of mass position to Å and comparing to given options

Now, we convert the center of mass position back to Å: CM \(= 1.2444 * 10^{-10} meters * \frac{1 Å}{10^{-10} meters}\) CM \(= 1.2444 Å\) Comparing this result with the given options: A) \(1 Å\) B) \(2.5 Å\) C) \(1.24 Å\) D) \(1.5 Å\) The correct choice is option C) \(1.24 Å\), as it is the closest to our calculated result of 1.2444 Å.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The centre of mass of a systems of two particles is (A) on the line joining them and midway between them (B) on the line joining them at a point whose distance from each particle is proportional to the square of the mass of that particle. (C) on the line joining them at a point whose distance from each particle inversely proportional to the mass of that particle. (D) On the line joining them at a point whose distance from each particle is proportional to the mass of that particle.

Two identical hollow spheres of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is \(\\{\mathrm{A}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(4 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(32 / 3) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(34 / 3) \mathrm{MR}^{2}\)

A circular disc of radius \(\mathrm{R}\) and thickness \(\mathrm{R} / 6\) has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and re-casted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is \(\ldots\) \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{\mathrm{B}\\}(2 \mathrm{I} / 8)\) \(\\{\mathrm{C}\\}(\mathrm{I} / 5)\) \(\\{\mathrm{D}\\}(\mathrm{I} / 10)\)

A mass \(\mathrm{m}\) is moving with a constant velocity along the line parallel to the \(\mathrm{x}\) -axis, away from the origin. Its angular momentum with respect to the origin \(\\{\mathrm{A}\\}\) Zero \\{B \\} remains constant \(\\{\mathrm{C}\\}\) goes on increasing \\{D\\} goes on decreasing

Two spheres each of mass \(\mathrm{M}\) and radius \(\mathrm{R} / 2\) are connected with a mass less rod of length \(2 \mathrm{R}\) as shown in figure. What will be moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod? \(\\{\mathrm{A}\\}(21 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(2 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(5 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(5 / 21) \mathrm{MR}^{2}\)
See all solutions

Recommended explanations on English Textbooks

Language Analysis

Read Explanation

Text Comparison

Read Explanation

Semiotics

Read Explanation

Textual Analysis

Read Explanation

International English

Read Explanation

English Grammar Summary

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free