In a $C O$ molecule, the distance between $C$ (mass $= 12$ a.m.u) and $O$ (mass $= 16$ a.m.u.), where 1 a.m.u $= \frac{5}{3} \times 10^{- 27} k g,$ is close to

2.4 \times 10^{- 10} m

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1.9 \times 10^{- 10} m

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1.3 \times 10^{- 10} m

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4.4 \times 10^{- 11} m

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Solution

The correct option is B $1.3 \times 10^{- 10} m$
$r_{1} = \frac{m_{2} d}{m_{1} + m_{2}}$ and $r_{2} = \frac{m_{1} d}{m_{1} + m_{2}}$
$I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2}$
$∴ d = 1.3 \times 10^{- 10} m$ .

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In a CO molecule, the distance between C (mass =12 a.m.u) and O (mass =16 a.m.u.), where 1 a.m.u =53×10−27kg, is close to

The correct option is B 1.3×10−10mr1=m2dm1+m2 and r2=m1dm1+m2I=m1r21+m2r22∴d=1.3×10−10m.

In a $C O$ molecule, the distance between $C$ (mass $= 12$ a.m.u) and $O$ (mass $= 16$ a.m.u.), where 1 a.m.u $= \frac{5}{3} \times 10^{- 27} k g,$ is close to

The correct option is B $1.3 \times 10^{- 10} m$
$r_{1} = \frac{m_{2} d}{m_{1} + m_{2}}$ and $r_{2} = \frac{m_{1} d}{m_{1} + m_{2}}$
$I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2}$
$∴ d = 1.3 \times 10^{- 10} m$ .