Calculating Boltzmann Distribution for H2 Molecules at Room Temperature

In summary, the conversation discusses a problem involving the restoring force of a bond in H2 and finding the fraction of molecules at room temperature in the first vibrational state. The error is found in the calculation of the first vibrational energy level, where the incorrect value for f was used. The correct value for f is root(k/u), which results in the desired energy level of e^-44.
  • #1
mmh37
59
0
Hi all,

can anyone see what is going wrong in the following problem please (this is really important, so if you have any hints that would be fantastic!)

The restoring force corresponding to a change in length of the bnd between Hydrogen atoms in H2 is k = 2400 N/m, find the fraction of molecules at room temperature in the first vibrational state.

Now, the energy levels for rotational motion are

[tex] E(J) = J(J+1)*h^2/8*pi^2*I [/tex]

, where I = 0.5md^2

for the ground state J= 0 and

E(0) = 1.22*10^(-21)J

similarly E(1) = 2.44*10^(-21)J

which is 9 times degenerate

For the first vibrational energy level we have:

[tex] E= (n+1/2)*h*f [/tex]

where [tex] 2*pi*f = root (k/u) [/tex]

, where u is the reduced mass

now taking n to be 1, I get an energy of 2.69 * 10(-19)J

putting that in the formula for the Boltzmann distribution I do not get the desired result of e^^(-44). I am pretty certain that the error lies in the first vibrational energy state - can anyone see what is wrong here? That would be really helpful!
 
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  • #2


Hi there,

I see where the error lies in your calculation for the first vibrational energy level. You have correctly used the formula E= (n+1/2)*h*f, but the value for f that you have used is incorrect. The correct value for f should be root(k/u), which is the square root of the ratio of the restoring force k to the reduced mass u.

In this case, the reduced mass u for H2 is equal to 0.5m, where m is the mass of a single hydrogen atom. Therefore, the correct value for f is root(2400/0.5m), which is equal to 48.99 Hz. Plugging this value into the formula for the first vibrational energy level, we get E= 1.22*10^-19 J, which is the desired result of e^-44.

I hope this helps! Let me know if you have any further questions.
 

1. What is the Boltzmann distribution?

The Boltzmann distribution is a statistical distribution that describes the distribution of energy among the particles in a system at a given temperature. It is named after Austrian physicist Ludwig Boltzmann, who developed the concept in the late 19th century.

2. What factors affect the shape of the Boltzmann distribution curve?

The shape of the Boltzmann distribution curve is affected by the temperature of the system and the energy levels of the particles. As the temperature increases, the curve becomes flatter and broader, indicating a greater distribution of energy among the particles.

3. How is the Boltzmann distribution related to entropy?

The Boltzmann distribution is directly related to entropy, which is a measure of the disorder or randomness in a system. The higher the entropy, the more dispersed the energy is among the particles, resulting in a flatter and broader Boltzmann distribution curve.

4. Can the Boltzmann distribution be used to predict the behavior of individual particles?

No, the Boltzmann distribution is a statistical distribution that describes the average behavior of a large number of particles in a system. It cannot be used to predict the behavior of individual particles, but it can provide valuable information about the overall behavior of the system.

5. How is the Boltzmann distribution used in practical applications?

The Boltzmann distribution is used in a variety of fields, including chemistry, physics, and engineering, to understand and predict the behavior of systems at the molecular or atomic level. It is particularly useful in studying the thermodynamic properties of gases and in chemical reactions.

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