FIGURE 10.23 showed the potential-energy curve for the O2 molecul...

Question

FIGURE 10.23 showed the potential-energy curve for the O2 molecule. Consider a molecule with the energy E1 shown in the figure.

a. What is the maximum speed of an oxygen atom as it oscillates back and forth? Don't forget that the kinetic energy is the total kinetic energy of the system. The mass of an oxygen atom is 16 u, where 1 u=1 atomic mass unit =1.66×10(to the poer of)−27 kg .

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to determine the maximum speed of oscillation of a nitrogen atom. If the minimum potential energy of the nitrogen atom is 1.79 times 10 to the power of minus joles. Uh we want to consider the mass of a nitrogen atom to be 14 U where one U is equals to one atomic mass unit which equals to 1.66 times 10 to the power of minus 27 kg. The maximum spit of oscillation of the nitrogen atom given R a 3.78 times 10 to the power of three m per second. B 5.56 times 10 to the power of three m per second. C 2.78 times 10 to the power of three m per second and D 6.56 times 10 to the power of three m per second. Awesome. So the way we want to tackle this problem is we will, we will use the particle model for each nitrogen atom. So the value of the kinetic energy or the maximum speed of oscillation will be able to be found when the potential energy is actually at its minimum value. So therefore, K max or the maximum kinetic energy will happen or will be equal to the minimum potential energy or human since there are two nitrogen atoms, so that will be two times K max will then equals to human 42 n nitrogen atoms just like. So, all right. So now we wanna actually substitute our equation for kinetic energy. We wanna recall that kinetic energy will be calculated by multiplying half multiplied by M multiplied by the velocity squared. So the left side of our equation will then be two multiplied by half, multiplied by M multiplied by V max squared. So V max is the maximum speed of oscillation which we are interested at, which is being asked by the problem statement that will equals to the minimum potential energy, which is going to be 1.79 times 10 to the power of minus 19 just like. So we wanna read arrange this so that we actually can find V max squared equals two. So the two and the half will cancel each other out. So that will leave us with 1.79 times 10 to the power of minus 19 jo divided by the mass. So V X squared will then be 1.17, 9 times 10 to the power of minus 19 Jule divided by the mass of the nitrogen atom, which is going to be 14 U or essentially going to be 1.79 times 10 to the power of minus 19 Jule divided by 14 U will equals to 14 multiplied the multiplied by a one atomic mins unit which is 1.66 times 10 to the power of minus kg, which is going to be 14 multiplied by 1.66 times 10 to the power of minus 27 kg and calculating. And then taking the square root of that, you will then get our V max value to then B 2.78 times 10 to the power of three m per second, which will correspond to answer C in our answer choices. So answer C with the maximum speed of oscillation of a nitrogen atom being 2.78 times 10 to the power of three m per second squared is going to be the answer to this particular practice problem. So if you guys still have any sort of confusion on this video, please make sure to check out our a lesson videos and I'll be all for this one. Thank you.

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