The probability for a diatomic molecule to occupy vibrational level n is given b

Question

The probability for a diatomic molecule to occupy vibrational level n is given by Pn = exp[-(n + 1/2) hv/kTJ/q_vib Derive an equation for the probability to find a diatomic molecule in its ground vibrational state, which has n = 0. You can use the result for q_vib from Lecture Notes 2. From your result in part a, you can deduce an expression for the probability that the molecule is in an excited vibrational state. What is the expression e., what is the sum of the probabilities that the molecule is in the state with n = 1, or n = 2, or n = 3, or...? For the 35 CI_2 molecule, hv/k = 805 K. Find a numerical value for the probability that the molecule is in its ground vibrational state (n = 0), when T = 1000 K. Derive the ratio between the probability that the 35Cl_2 molecule is in the vibrational state with quantum number n + 1, relative to the probability that it is in the vibrational state with quantum number n, at an arbitrary temperature T. Using your results from parts c and d, compute the probability for the 35CI_2 molecule to be in the vibrational states with quantum numbers n = 1, n = 2, n = 3, .. at T = 1000 K. Find a separate value for each state. Continue to increase n and compute the probability Pn until pm becomes extremely small. Then add the probabilities for have n = 0, n = 1, n = 2, n = 3 ... and check that the total comes very close to 1, at T = 1000 K.

Explanation / Answer

ANS:

A) Pn= exp{-(n+1/2)hv/kt} / qvib

for ground vibrational level, n=0

and for diatomic molecules, qvib = (1/ 1-e-hv/kt)

Thus deriving the equation of probability for diatomic molecule in ground vibrational state is:

P0 = exp [ -1/2 hv/kt] / qvib

     = exp [ -1/2 hv/kt] *(1-e-hv/kt )

solving this , we get

P0 = e-1/2 hv/kt -e-3/2 hv/kt

P0 = e-2hv/kt

B)

now, P0 =e-1/2 hv/kt / qvib

            P1 = e-3/2 hv/kt / qvib

        .

        .

        Pn= e-(n+1/2) hv/kt

Adding all the above equations, we get

P0 +P1 + P2...... +Pn = ( e-1/2 hv/kt + e-3/2 hv/kt ......+e-(n+1/2) hv/kt )/ qvib

                                              = e-1/2 hv/kt (1 + e-hv/kt + e-2hv/kt +.....) / qvib

C) for 35Cl2 , hv/k= 805

and t=1000 k

thus P0 = e -2*805/ 1000

             = e -1.610

Thus P0 = 5.002.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.