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Question

If the distance between the centers of the atoms of potassium and bromine in $${\text{KBr}}$$ (potassium-bromide) molecule is $$0.282 \times {10^{ - 9}}m,$$ find the centre of mass of this two-particle system from potassium (mass of bromine$$ = 80\,u,$$ and of potassium $$ = 39\,u$$ ).


Solution

Let position coordinate of potassium,$$X_K=0$$

Position co-ordinate of bromine,
$$X_{Br}=0.282\times 10^{-9}m$$.

Position co-ordinate of centre of mass-
$$X_{CM}= \dfrac{m_kX_k+m_{Br}X_{Br}}{m_k+m_{Br}}$$

$$X_{CM}=\dfrac{39\times 0+80\times 0.282\times 10^{-9}}{39+80}$$
 
$$=0.189\times 10^{-9}m$$

Chemistry

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