1 Molecular Geometry (a) Electron diffraction experiments reveal that the distan

Question

1 Molecular Geometry (a) Electron diffraction experiments reveal that the distance between the centers of the carbon (C) and oxygen (O) atoms in a carbon monoxide molecule is about 1.100 10-10 m. Where is the center of mass of the CO molecule, relative to the carbon atom? (b) In an ammonia molecule (NH3), the three hydrogen atoms (H) form an equilateral triangle at the base of a pyramid. The distance between the centers of the hydrogen atoms is 1.628-10-10 m. Thus, the center of the triangle is 9.399 10-10 m from each hydrogen atom. The nitrogen atom (N) sits at the apex of the pyramid. The distance between the center of the nitrogen atom and any hydrogen atom is 1.014 -10-10 m Locate the center of mass of the ammonia molecule, relative to the nitrogen atom (The nitrogen atom actually oscillates up and down through the base of the pyramid. For this problem, assume a static molecule.) A radioactive nucleus, originally at rest, decays by emitting an electron and a neutrino at right angles to one another. The momentum of the electron is 1.2. 10-25 kg m/s and that of the neutrino is 1.6 10-2 kg m/s. (a) Find is the momentum of the recoiling nucleus. Report both the magnitude of its momentum and its direction relative to the electron. (b) If the speed of the residual nucleus is 3.3 102 m/s, what is its mass? 3 A Speeding Bullet A bullet with a mass of 5.0 g is fired horizontally into a 2.0 kg wooden block that is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.20. After the bullet strikes the block. the block slides along the surface. It comes to rest after it moves 2.0 m What was the initial speed of the bullet?

Explanation / Answer

mass of carbon atom=12 amu

mass of oxygen atom=16 amu

let carbon atom be at origin and oxygen be at a distance of 1.1*10^(-10) m

then center of mass from origin=(12*0+16*1.1*10^(-10))/(12+16)=0.68257*10^(-10) m

part b:

let line connecting center of the triangle and the nitrogen atom has a length of d.

then considering the right angle triangle made by this line, line connecting the nitrogen atom and the hydrogen atom and line connecting the hydrogen atom to the center of the triangle,

d^2+(9..399*10^(-11))^2=(1.014*10^(-10))^2

==>d=sqrt(1.014^2-0.9399^2)*10^(-10)=0.3805*10^(-10) m

then center of mass from the nitrogen atom=(14*0+3*1*0.3805*10^(-10))/(14+3*1)

=6.7147*10^(-12) m

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