1. (This problem is similar to Problem 5-2 in McQuarrie and Simon) Make the \"ha
ID: 707065 • Letter: 1
Question
1. (This problem is similar to Problem 5-2 in McQuarrie and Simon) Make the "harmonic approximation" to the following diatomic potential energy function to obtain the corresponding harmonic potential energy function of the form VHo (R)R- R as illustrated in the figure below. Find expressions for the harmonic force constant k and equilibrium internuclear distance R in terms of the parameters of the potential function, and clearly identify them as your answers The potential is the Lennard-Jones Potential 12 L-J This potential is used to approximate weak attractive interactions between atoms that are not covalently bonded, like Ar2. Hint: start by showing that the minimum of the potential function occurs at an internuclear distance of 216? VL-J(R) 2.0 1.5 0.5 0 2 4Explanation / Answer
A simple function that is frequently used to describe the potential energy of rare gas dimers (such as Ar2) is the Lennard-Jones potential
VL-J(R) = 4d((?/R)12-( ?/R)6) + d
where R is the distance between the two atoms and the two parameters, d and ? depend on which atoms are involved. The shape of the curve is qualitatively similar to the Morse potential curve, but is more appropriate for van der Waals interactions (while Morse is more appropriate for molecules with a covalent bond between atoms). In each one of the questions (a-e) below you should give an expression that contains the parameters of the potential function, d and ?, and possibly also the mass of the two atoms, m1 and m2.
At the minimum, the first derivative is zero.
Find dV/dR by differentiating the expression for the Lennard-Jones potential, and then set it to zero to find Rb.
dV/dR = 4d ( ?12( ? 12 /R 13 )+ 6 (? 6 /R 7) = 24d/ R( ?2 (?/ R)12 + (?/ R)6 )
This must be zero when R = Rb, so Rb must satisfy
2 (?/ R)12 = (?/ R)6
That is
Rb = 2 1/6 ?
The potential energy at the minimum is V(Rb)
V(Rb) = 4d(( ?/ 2 1/6 ? )12 –( ?/ 2 1/6 ? )6) = 4d( 1/ 4 – 1/ 2) = ?d
Let x = R-Rb and V(x) = V(R). Need to expand V(x) up to and including the second order term in a Taylor series
V(x) = V(0) + x(dV/dx(0)) + x2/2(d2V/dx2(0))
= -d + 0+(1/2)kx2
Where k = d2V/dx2(0). The first derivative term vanishes because x = 0 corresponds to a minimum of the V(x) function. This is a potential energy function for a harmonic oscillator with a spring constant k. In oder to get an expression for the spring constant in terms of the potential parameters d and ? and the atomic masses m1 and m2, find the second derivative
k= d2V/dx2(0) = d2V/dR2(Rb) = 4d ( 12*13( ? 12 /Rb14 )- 6*7 (? 6 /Rb8)
= (4d/Rb2) (12*13( ? /Rb )12- 6*7 (? /Rb )6)
= (4d/(21/6 ?)2) (12*13( 1 /2)2- 6*7 (1/2 ))
= (4d/(21/3 ?2))(39-21)
= 72d/(21/3 ?2)
= 36d/(22/3 ?2)
The dissociation energy is the energy needed to bring the dimer up from the ground state to the state of two separated atoms.
Ed = E2AR – EAR2
The energy of two separated atoms is E2AR = lim R->infinity = 0 and the energy of the ground state is
EAR2 = -d+(h/2) ?(k/µ)
where k is given by the expression obtained above and µ is the reduced mass of the dimer µ = m1m2/(m1 + m2). It is important to add to ?d the zero point energy of the harmonic oscillator, which is the energy of the ground state with respect to the potential energy minimum.
The energy required to excite a harmonic oscillator from a stationary state to the next higher state is
?E = h*w = h ?(k/µ)
Where the expression for k and µ are given in the above parts.
The first anharmonic correction term to the harmonic oscillator approximation to the Lennard-Jones potential is the third order term in the Taylor expansion, the next term after the second order expansion given in part above part. This is
?E = (x3/3!) (d3V/DR3)(Rb)
The expression for the third derivative can be obtained by differentiating the second derivative given in part
d3V/DR3(Rb) = 4d ( -12*13*14( ? 12 /Rb15 )- 6*7*8 (? 6 /Rb9)
Now plug in Rb = 21/6 and simplify.
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