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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.

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Most popular questions from this chapter

A cylinder of mass \(5 \mathrm{~kg}\) and radius \(30 \mathrm{~cm}\), and free to rotate about its axis, receives an angular impulse of $3 \mathrm{~kg} \mathrm{M}^{2} \mathrm{~S}^{-1}$ initially followed by a similar impulse after every \(4 \mathrm{sec}\). what is the angular speed of the cylinder 30 sec after initial impulse? The cylinder is at rest initially. \(\\{\mathrm{A}\\} 106.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{B\\} \(206.7 \mathrm{rad} \mathrm{S}^{-1}\) \\{C\\} \(107.6 \mathrm{rad} \mathrm{S}^{-1}\) \\{D \(\\} 207.6 \mathrm{rad} \mathrm{S}^{-1}\)

Consider a two-particle system with the particles having masses \(\mathrm{M}_{1}\), and \(\mathrm{M}_{2}\). If the first particle is pushed towards the centre of mass through a distance \(d\), by what distance should the second particle be moved so as to keep the centre of mass at the same position? $\\{\mathrm{A}\\}\left[\left(\mathrm{M}_{1} \mathrm{~d}\right) /\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)\right]$ $\\{\mathrm{B}\\}\left[\left(\mathrm{M}_{2} \mathrm{~d}\right) /\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)\right]$ $\\{\mathrm{C}\\}\left[\left(\mathrm{M}_{1} \mathrm{~d}\right) /\left(\mathrm{M}_{2}\right)\right]$ $\\{\mathrm{D}\\}\left[\left(\mathrm{M}_{2} \mathrm{~d}\right) /\left(\mathrm{M}_{1}\right)\right]$

A uniform rod of length \(\mathrm{L}\) is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length, what maximum speed must be imparted to the lower end so that the rod completes one full revolution? \(\\{\mathrm{A}\\} \sqrt{(2 \mathrm{~g} \mathrm{~L})}\) \(\\{\mathrm{B}\\} 2 \sqrt{(\mathrm{gL})}\) \(\\{C\\} \sqrt{(6 g L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 g \mathrm{~L})}\)

Three particles of the same mass lie in the \((\mathrm{X}, \mathrm{Y})\) plane, The \((X, Y)\) coordinates of their positions are \((1,1),(2,2)\) and \((3,3)\) respectively. The \((X, Y)\) coordinates of the centre of mass are \(\\{\mathrm{A}\\}(1,2)\) \(\\{\mathrm{B}\\}(2,2)\) \(\\{\mathrm{C}\\}(1.5,2)\) \(\\{\mathrm{D}\\}(2,1.5)\)

A binary star consist of two stars \(\mathrm{A}(2.2 \mathrm{Ms})\) and \(\mathrm{B}\) (mass \(11 \mathrm{Ms}\) ) where \(\mathrm{Ms}\) is the mass of sun. They are separated by distance \(\mathrm{d}\) and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of \(\operatorname{star} B\). about the centre of mass is \(\\{\mathrm{A}\\} 6\) \(\\{\mathrm{B}\\} \overline{(1 / 4)}\) \(\\{C\\} 12\) \(\\{\mathrm{D}\\}(1 / 2)\)
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