Question

If the distance between the centres of the atoms
of potassium and bromine in KBr molecule is
0.282 x 10 m, the centre of mass of this two
particle system from potassium is (mass of
bromine = 80 u and of potassium = 39 u).
1) 9.3 x 10-2 nm

2) 18.96 102 nm

3) 27.9 x 10-2 nm

4) 3.3 * 10-2 nm
​

Answer 1

xcm=m1x1+m2x2

.................

Answer 2

The position coordinate of center of mass is $18.96\times10^{-2}\ nm$

(2) is correct option.

Explanation:

Given that,

Mass of bromine = 80 u

Mass of potassium = 39 u

Position co-ordinate of bromine $d=0.282\times10^{-9}\ m$

Let the position coordinate of potassium is zero.

We need to calculate the position coordinate of center of mass

Using formula of center of mass

$X_{cm}=\dfrac{m_{k}X_{k}+m_{br}X_{br}}{m_{k}+m_{br}}$

Put the value into the formula

$X_{cm}=\dfrac{39\times0+80\times0.282\times10^{-9}}{39+80}$

$X_{cm}=18.96\times10^{-11}\m$

$X_{cm}=18.96\times10^{-2}\ nm$

Hence, The position coordinate of center of mass is $18.96\times10^{-2}\ nm$

Learn more :

Topic : center of mass

https://brainly.in/question/799070