The water molecule consists of an oxygen atom (mass 16 u) and two hydrogen atoms
ID: 2077852 • Letter: T
Question
The water molecule consists of an oxygen atom (mass 16 u) and two hydrogen atoms (each mass 1 u), with the equilibrium bond length from hydrogen to oxygen of 95.9 pm and an angle of 104.5° between the two hydrogen bonds. (A) Where is the center of mass of the molecule, and what is the moment of inertia of the molecule about an axis through the center of mass running perpendicular to the plane of the molecule? You may ignore the electrons' contribution to the masses, and you may treat the atomic nuclei as points (the atomic nucleus is approximately 1 fm in size). (B) The same as for part (A), but with both hydrogens replaced by deuterium atoms (mass 2 u).Answer all parts thank you .
The water molecule consists of an oxygen atom (mass 16 u) and two hydrogen atoms (each mass 1 u), with the equilibrium bond length from hydrogen to oxygen of 95.9 pm and an angle of 104.5° between the two hydrogen bonds. (A) Where is the center of mass of the molecule, and what is the moment of inertia of the molecule about an axis through the center of mass running perpendicular to the plane of the molecule? You may ignore the electrons' contribution to the masses, and you may treat the atomic nuclei as points (the atomic nucleus is approximately 1 fm in size). (B) The same as for part (A), but with both hydrogens replaced by deuterium atoms (mass 2 u).
Answer all parts thank you .
The water molecule consists of an oxygen atom (mass 16 u) and two hydrogen atoms (each mass 1 u), with the equilibrium bond length from hydrogen to oxygen of 95.9 pm and an angle of 104.5° between the two hydrogen bonds. (A) Where is the center of mass of the molecule, and what is the moment of inertia of the molecule about an axis through the center of mass running perpendicular to the plane of the molecule? You may ignore the electrons' contribution to the masses, and you may treat the atomic nuclei as points (the atomic nucleus is approximately 1 fm in size). (B) The same as for part (A), but with both hydrogens replaced by deuterium atoms (mass 2 u). You may ignore the electrons' contribution to the masses, and you may treat the atomic nuclei as points (the atomic nucleus is approximately 1 fm in size). (B) The same as for part (A), but with both hydrogens replaced by deuterium atoms (mass 2 u).
Answer all parts thank you .
Explanation / Answer
a)
Taking oxygen atom as origin and hydrogen atoms placed symmetrically about x-axis, let's calculate the center of mass and moment of inertia
For x-axis
M * rcx = Mo*rox + Mh * (rhx - rhx)
For y-axis
M * rcy = Mo * roy + Mh * ( rhy + rhy )
rox = roy = 0
rhx = rh * sin t = 75.83 pm where t = 104.5/2 and ro = 95.9 pm
rhy = rh * cos t = -58.71 pm
Total Mass = (16 + 2)u = 18 u
=> rcx = 0
=> rcy = (0 - 2*1*58.71)/18 = -6.52 pm
Moment of Inertia I = sum over i(Mi * ri^2) where ri = distance from center of mass
=> I = 16 * 6.52^2 + 2 * ((58.71 - 6.52)^2 + 75.83^2) = 17628 (u * pm^2)
=> I = 1.76 * 10^4 (u * pm^2)
b) Taking oxygen atom as origin and deuterium atoms placed symmetrically about x-axis, let's calculate the center of mass and moment of inertia
For x-axis
M * rcx = Mo*rox + Md * (rdx - rdx)
For y-axis
M * rcy = Mo * roy + Md * ( rdy + rdy )
rox = roy = 0
rdx = rd * sin t = 75.83 pm where t = 104.5/2 and ro = 95.9 pm
rdy = rd * cos t = -58.71 pm
Total Mass = (16 + 4)u = 20 u
=> rcx = 0
=> rcy = (0 - 2*2*58.71)/20 = -11.74 pm
Moment of Inertia I = sum over i(Mi * ri^2) where ri = distance from center of mass
=> I = 16 * 11.74^2 + 4 * ((58.71 - 11.74)^2 + 75.83^2) = 34031 (u * pm^2)
=> I = 3.40 * 10^4 (u * pm^2)
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