If the distance between the centres of the atoms of potassium and bromine in KBr molecule is $$0.4$$×$$10$$−$$9$$ m, the centre of mass of this two particle system from potassium is (mass$$ of bromine $$=$$ $$80$$ u and of potassium $$=$$ $$39$$ $$u).

Correct option is 226×10−2 nm

Questions

If the distance between the centres of the atoms of potassium and bromine in KBr molecule is $0. 4 \times 1 0^{- 9} m,$ the 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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9.3×10−2 nm

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detailed solution

Correct option is B

x=m2dm1+m2=80×0.4×10−980+39 =32119×10−9 =0.26×10−9 =26×10−2nm

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If the distance between the centres of the atoms of potassium and bromine in KBr molecule is 0.4×10−9 m, the 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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If the distance between the centres of the atoms of potassium and bromine in KBr molecule is $0. 4 \times 1 0^{- 9} m,$ the centre of mass of this two particle system from potassium is (mass of bromine = 80 u and of potassium = 39 u).