7.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

11 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Let us assume that H atom and Cl atom are at a distance of x and y respectively from the CM (Centre of Mass).

If mass of the H atom = m, mass of the Cl atom = 35.5m

Given x + y = 1,27 À

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore,

We can have, : (my+35.5mx)/ (m+35.5m) = 0

mx + 35.5my = 0

x = 35.5 (1.27 – x)

x = 1.24 À

So the centre of mass lies 1.24 À from H atom

0 Upvotes 0 Downvotes

Similar Questions for you

The position vector of 1kg object is $\vec{r} = (3 \hat{i} - \hat{j}) m$ and its velocity $\vec{v} = (3 \hat{j} + \hat{k}) m s^{- 1} .$ The magnitude of is angular momentum is $\sqrt[]{x} N m$ where x is________.

Raj Pandey

$\overset{?}{L} = \overset{?}{r} * m \overset{?}{v}$

= $(3 \hat{i} ? \hat{j}) * (3 \hat{j} + \hat{k})$

$= 9 \hat{k} + 3 (? \hat{j}) ? \hat{i}$

$= ? \hat{i} ? 3 \hat{j} + 9 \hat{k}$

$| \overset{?}{L} | = \sqrt[]{1 + 9 + 81} = \sqrt[]{91}$

Match List – I with List – II

Choose the correct answer from the options given below:

Raj Pandey

Use formula for M.I.

A shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio $2 : 2 : 1$ . If the fragments having equal mass fly off along mutually perpendicular directions with speed $v$ , the speed of the third (lighter) fragment is :

Raj Pandey

The angular speed of a fly wheel moving with uniform angular acceleration changes from 1200rpm to 3120 in 16 seconds. The angular acceleration in $r a d / s^{2}$ is:

Raj Pandey

$\begin{array}{r} ω = ω_{0} + α t \\ α = \frac{ω - ω_{0}}{t} \\ = \frac{(3120 - 1200)}{16 s} r p m \\ = \frac{1920}{16} \times \frac{2 π}{60} r a d / s^{2} \\ = 4 π r a d / s^{2} \end{array}$

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is :

Raj Pandey