In a CO molecule, the distance between C (mass = 12 amu) and O (mass = 16 amu), where 1 amu =(5/3)×10-27 kg, is close to

Question

In a CO molecule, the distance between C (mass = 12 amu) and O (mass = 16 amu), where 1 amu =(5/3)×10-27 kg, is close to (A) 2.4×10-10m (B) 1.9×10-10m (C) 1.3×10-10m (D) 4.4×10-10m

Tardigrade Admin · Accepted Answer

I=μ r2 (where μ=reduced mass) μ=(m1m2/m1+m2)=(48/7)amu =11.43×10-27kg Substituting in I=μ r2 we get, √(1.87×10-46/11.43×10-27) =1.28×10-10m ∴ The correct answer is (c).

Q. In a CO molecule, the distance between C (mass = 12 amu) and O (mass = 16 amu), where 1 amu $= \frac{5}{3} \times 1 0^{- 27}$ kg, is close to

$2.4 \times 1 0^{- 10} m$

$1.9 \times 1 0^{- 10} m$

$1.3 \times 1 0^{- 10} m$

$4.4 \times 1 0^{- 10} m$

$I = μ r^{2}$ (where $μ$ =reduced mass)
$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{48}{7} am u$
$= 11.43 \times 1 0^{- 27} k g$
Substituting in $I = μ r^{2}$ we get,
$\frac{1.87 \times 1 0 ^{- 46}}{11.43 \times 1 0 ^{- 27}}$
$= 1.28 \times 1 0^{- 10} m$
$∴$ The correct answer is (c).

Q. In a CO molecule, the distance between C (mass = 12 amu) and O (mass = 16 amu), where 1 amu =35​×10−27 kg, is close to

2.4×10−10m

1.9×10−10m

1.3×10−10m

4.4×10−10m