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If the distance between the centres of the atoms of potassium and bromine in KBr molecule is 0.282 × 10^-9 m, the distance of centre of mass of this two particle system from potassium is (mass of bromine = 80 u and of potassium = 39 u).

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18.96 × 10^-2 nm

9.3 × 10^-2 nm

27.9 × 10^-2 nm

3.3 × 10^-2 nm

answer is A.

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Detailed Solution

Distance of CM from m₁ is

$= \frac{m_{2} d}{m_{1} + m_{2}} = \frac{80 \times 0.282 \times 10^{- 9}}{39 + 80} = 0.1896 \times 10^{- 9} m = 18.96 \times 10^{- 2} nm$

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