[PoM] Lennard-Jones potential parameters

[BRN]

Hi guys! I need your help!

1. Homework Statement

Use the function of Lennard-Jones V (R) = ε [(σ / R)¹² - (σ / R)⁶] as model for the adiabatic potential energy in function of the separation between the ¹¹B boron and nitrogen nuclei ¹⁴N. You determine the parameters ε and σ to reproduce the spectroscopic values of the vibrational quantum, ν = 1514.6 cm^-1, and the separation of 1,666 cm^-1 between the lines of rotational BN molecule.

The Attempt at a Solution

[/B]
I have

ΔE_rot=ħ²/I=1.666 cm^-1=3.3091*10^-23 J ⇒ I=ħ²/ΔE_rot=3.3603*10^-46 Kgm²

and

R₀=√(I/μ)=1.8125*10^-10 m

At this point, I calculating the minimum oh the potential:

V(0)=∂V(R)/∂R =0 ⇒ R₀⁶=2σ⁶ ⇒ σ=⁶√(R₀⁶/2)=1.6147*10^-10 m

For ε, I calculating k by:

k=∂²V(R)/∂R²=6σ⁶εR^-14(26σ⁶-7R⁶)

but here I'm lost...

Who can help me? Please!

 

[DrClaude]

For ε, I calculating k by:

k=∂²V(R)/∂R²=6σ⁶εR^-14(26σ⁶-7R⁶)

but here I'm lost...

You need an equation to relate k to $\nu$, then the only unknown in there will be ε.

 

[BRN]

Tanks for your answer,
the only equation i know to relate k to ν is:

ν=1/(2πc)√(k/μ)

so, i have (using R=R₀):

ε=[μ(2πcν)²σ⁷(⁶√2)¹⁴]/(72σ⁶)

but is dimensional incorrect...

 

[TSny]

For ε, I calculating k by:

k=∂²V(R)/∂R²=6σ⁶εR^-14(26σ⁶-7R⁶)

When you let R = R_o, you should then be able to simplify to get a nice expression relating $k$ to $\varepsilon$ and $\sigma$. This might make it a little easier to get the correct expression for $\varepsilon$ in terms of $\nu$.

 

[TSny]

Tanks for your answer,
the only equation i know to relate k to ν is:

ν=1/(2πc)√(k/μ)

so, i have (using R=R₀):

ε=[μ(2πcν)²σ⁷(⁶√2)¹⁴]/(72σ⁶)

but is dimensional incorrect...

I think your expression for $\varepsilon$ is correct except that you appear to be off by a factor of $\sigma$. Try to simplify the expression as much as possible.

 

[BRN]

Oh Damn!
yes, the equation is correct (apart the simplifications), but υ must be convert to m^-1 and not to J!

So:

ε=[μ(2πcν)²σ²(6√2)¹⁴]/72=1.5192*10^-18 J

Now it's ok!

Tanks at all!