What Is the Interatomic Distance at Which V(r) Is Minimized in Cl2?

[ibysaiyan]

Homework Statement

The interatomic force in the chlorine molecule Cl2 may be represented
by the Lennard-Jones potential:

with e = 1.79.10-19 J and r0 = 0.2 nm.

(i) Find the interatomic distance at which V(r) is minimized. What is the interatomic
force at this separation?
(ii) Calculate the potential energy at this separation.
(iii) Write down an approximate form for the potential at larger separations.
Comment (quantitatively) on how good this approximation is.
(iv) Make a simple estimate of the temperature required to cause Cl2 molecules to
dissociate.
(v) What is the frequency of vibration of the bond length?

Homework Equations

V(r) = 4 \epsilon [(\sigma/r)^12 ]- [ (\sigma/r)^6]

 

[ibysaiyan]

My attempt:

The Attempt at a Solution

(i) Find the interatomic distance at which V(r) is minimized. What is the interatomic
force at this separation?

i) Now the way I approached this part is as following:
One of the point at which the potential is at minimum ( i.e V (r) = 0) is when the x-axis is at r = σ. So by algebraic manipulation and differentiation of the LJ expression I end up with : r_{0} = (2^1/6 )*σ

After plugging in the given data I end up with a distance of : 0.178nm

That answers the first bit... now it asks for a value of force.. but wouldn't the force be zero ? since F (r) = - du/dr ... ?
In the above scenario we have V(r) = 0 , so surely differentiating it will give us dv/dr = 0. ( I thought of this intuitively, I hope it's right)

(ii) Calculate the potential energy at this separation.
Part ii)
To calculate the P.e energy I simply plug in all the numbers into the LJ equation.. so i end up with ~ -2.17*10^-18 j .
(iii) Write down an approximate form for the potential at larger separations.Comment (quantitatively) on how good this approximation is.

part iii)
Well at as separation distance from equilibirum increases, force due to attraction takes the part until a point comes when the potential energy goes to zero as r -> ∞ ,now I am aware that the expression r^-6 is to do with attractive forces/ van der wall forces,etc..so how do I express this equation.. I mean doesn't larger distance imply r -> ∞ so the denominator of equation goes infinite.. so we get zero too ? or maybe I am overseeing things.EDIT: Or here's a thought would I simply ignore the r^-12 term since it's related to repulsion forces... hmm

Thanks for your help, feedback is much appreciated! :)

 

[ibysaiyan]

can anyone please go through my working ?

If there's something you don't understand I will try to elaborate further.. let me know.

 

[ibysaiyan]

I think part iii of the question simply wants an approximation of the potential so I suppose in this case I could simply ignore the inverse power 12 expression since that's to do with repulsion followed on by the comment that at greater separation there's less probability for the hard spheres to interact due to which V(r) = 0.

Part iv) it asks me to find the temperature.. now I have though a way or two of doing this.. i could try by using specific heat at constant volume formula or by using equipartition theorem , I have tried both which give me absurd value..

Any suggestions ?

 

[ehild]

The force is zero at the minimum of the potential .

At longer distances you can ignore the power 12 term.

You can approximate the potential around the equilibrium position as that of a spring, -kx². Using the reduced mass of the two chlorine atom, you can get the vibration frequency.

The energy of the chlorine molecule in equilibrium is the minimum potential energy. When the atoms are separated, their energy is at least zero. So the needed dissociation energy is equal to the minimum potential energy, with opposite sign.

You know from the equipartition principle, that 1/2K_BT energy belongs to every degrees of freedom. Here the vibration of the chlorine molecule counts, and the vibration has two degrees of freedom, E=k_BT, and this has to be equal to the dissociation energy.

ehild

 

[vela]

My attempt:

The Attempt at a Solution

(i) Find the interatomic distance at which V(r) is minimized. What is the interatomic
force at this separation?

i) Now the way I approached this part is as following:
One of the point at which the potential is at minimum ( i.e V (r) = 0) is when the x-axis is at r = σ. So by algebraic manipulation and differentiation of the LJ expression I end up with : r_{0} = (2^1/6 )*σ

After plugging in the given data I end up with a distance of : 0.178nm

That answers the first bit... now it asks for a value of force.. but wouldn't the force be zero ? since F (r) = - du/dr ... ?
In the above scenario we have V(r) = 0 , so surely differentiating it will give us dv/dr = 0. ( I thought of this intuitively, I hope it's right)

That's not correct. Why would V(r)=0 imply dV/dr=0? You want to minimize V(r). What did you learn in calculus class about finding the minimum of a function?

(ii) Calculate the potential energy at this separation.
Part ii)
To calculate the P.e energy I simply plug in all the numbers into the LJ equation.. so i end up with ~ -2.17*10^-18 j .
(iii) Write down an approximate form for the potential at larger separations.Comment (quantitatively) on how good this approximation is.

part iii)
Well at as separation distance from equilibirum increases, force due to attraction takes the part until a point comes when the potential energy goes to zero as r -> ∞ ,now I am aware that the expression r^-6 is to do with attractive forces/ van der wall forces,etc..so how do I express this equation.. I mean doesn't larger distance imply r -> ∞ so the denominator of equation goes infinite.. so we get zero too ? or maybe I am overseeing things.


EDIT: Or here's a thought would I simply ignore the r^-12 term since it's related to repulsion forces... hmm

Thanks for your help, feedback is much appreciated! :)

Once you find the minimum, expand V(r) in a Taylor series about the minimum.

 

[ibysaiyan]

The force is zero at the minimum of the potential .

At longer distances you can ignore the power 12 term.

You can approximate the potential around the equilibrium position as that of a spring, -kx². Using the reduced mass of the two chlorine atom, you can get the vibration frequency.

The energy of the chlorine molecule in equilibrium is the minimum potential energy. When the atoms are separated, their energy is at least zero. So the needed dissociation energy is equal to the minimum potential energy, with opposite sign.

You know from the equipartition principle, that 1/2K_BT energy belongs to every degrees of freedom. Here the vibration of the chlorine molecule counts, and the vibration has two degrees of freedom, E=k_BT, and this has to be equal to the dissociation energy.

ehild

The force is zero at the minimum of the potential .

At longer distances you can ignore the power 12 term.
Yes. I understand the above results.
You can approximate the potential around the equilibrium position as that of a spring, -kx². Using the reduced mass of the two chlorine atom, you can get the vibration frequency.
Oh so I can find temperature by letting epsilon equate to KbT .. I was on the right track on this however I wasn't sure whether to have 5 dof for the cl2 molecule and since total energy = f/2 Nkbt.. I had plugged N , which's number of particles as 2... to give out an improbable answer
I know that for a spring P.E is 1/2 kx^2 .. in this case we have two molecules vibrating so would the overall expression be 1kx^2 ?.. Oh I think i get it now.. first I find out the total P.E value and then let -kx^2 = P.E value... so I can get my spring constant because my 'x' i.e extension value will be the difference from equilibrium position to position 'r' also time period of oscillation = Square root of k/m hm... that makes sense however for reduced mass do I just plug in the g/mol value divided by Na of cl2 o I initially tried to approach this problem by assuming that 1/2 Iw^2 is the rotational kinetic energy ( conservation of energy, if that even makes sense) ... and some how get the value for moment of inertia to give to finally give me the angular velocity. Then I can simply divide 'w'/ 2pi for frequency...

That's not correct. Why would V(r)=0 imply dV/dr=0? You want to minimize V(r). What did you learn in calculus class about finding the minimum of a function?Once you find the minimum, expand V(r) in a Taylor series about the minimum.

Well at minimum the derivative is zero.. so I could differentiate the above equation wrt 'r' and we know that r = r0 , at minimum potential which's how I ended up with r0 = σ * 2^1/6
to give me the sigma value which's the inter atomic separation when V(r) = 0 ...

I don't know what is causing confusion here ( to me)... hm...

EDIT: Thanks for your replies.. I will definitely respond back soon. Maybe it's the lack of sleep due to which some bits of this question don't seem obvious to me.

 

[vela]

Where are you getting V(r)=0 from?

 

[ibysaiyan]

Where are you getting V(r)=0 from?

In my lecture notes there seems to be a condition placed that when V (r) =0 , r= sigma
Edit: I think that doesn't make any sense how about DV/Dr=0 , then whatever value I get can be plugged into second derivative to get my minimum

 

[vela]

OK, you said V'(r₀)=0 and solved for σ. So far, so good. Note that the force F(r)=-V'(r), so at the minimum of the potential, F=0, as ehild said above.

You have V(σ)=0, but that has nothing to do with the minimum. I think that's what confusing you. After all, in (ii), you found V(r₀)≠0.

What you want to do is expand V(r) as a Taylor series about r=r₀:
$$V(r) = V(r_0) + V'(r_0)(r-r_0) + \frac{V''(r_0)}{2!}(r-r_0)^2 + \frac{V'''(r_0)}{3!}(r-r_0)^3 + \cdots$$The idea is to truncate the series at the quadratic term, so you have a harmonic-oscillator potential. You already calculated the constant term. The first-order term drops out because V'(r₀)=0. The second-order term is the 1/2 kx² term you need.

The size of the third-order term gives you an idea of how good the approximation is. If it's small, you can infer that truncating the series before it will yield a good approximation.

 

[ibysaiyan]

OK, you said V'(r₀)=0 and solved for σ. So far, so good. Note that the force F(r)=-V'(r), so at the minimum of the potential, F=0, as ehild said above.

You have V(σ)=0, but that has nothing to do with the minimum. I think that's what confusing you. After all, in (ii), you found V(r₀)≠0.

What you want to do is expand V(r) as a Taylor series about r=r₀:
$$V(r) = V(r_0) + V'(r_0)(r-r_0) + \frac{V''(r_0)}{2!}(r-r_0)^2 + \frac{V'''(r_0)}{3!}(r-r_0)^3 + \cdots$$The idea is to truncate the series at the quadratic term, so you have a harmonic-oscillator potential. You already calculated the constant term. The first-order term drops out because V'(r₀)=0. The second-order term is the 1/2 kx² term you need.

The size of the third-order term gives you an idea of how good the approximation is. If it's small, you can infer that truncating the series before it will yield a good approximation.

Hi , sorry for my delayed reply. I was busy but now as I was just going through my work I have finally realized my mistake, that dR(r)/dr = 0 , doesn't imply V(r) = 0, since we can get maximum and minimums anywhere!

Also thanks for showing out on how to approach approximation for the particular question. That's very ingenious way I failed to realize that at separation distances closer to equilibrium LJ potential acts like a spring mass system.

I suppose I was trying my luck into solving this problem whilst having slept for maybe 3 hours but now each step shown to me makes sense. This is what I love about physics.


Also it's obvious now ( as was shown to me previously by one the respective members ) that to break apart the bonding of the two spheres the energy needed is the same energy, one which's found at the minimum potential depth with the difference being positive since we're going in +ve direction of the x-axis... to reach at a point where V(r) becomes zero as r -> infinite , beautiful.

I am working through part iv and v as I type this.

 

[ibysaiyan]

The force is zero at the minimum of the potential .

At longer distances you can ignore the power 12 term.

You can approximate the potential around the equilibrium position as that of a spring, -kx². Using the reduced mass of the two chlorine atom, you can get the vibration frequency.

The energy of the chlorine molecule in equilibrium is the minimum potential energy. When the atoms are separated, their energy is at least zero. So the needed dissociation energy is equal to the minimum potential energy, with opposite sign.

You know from the equipartition principle, that 1/2K_BT energy belongs to every degrees of freedom. Here the vibration of the chlorine molecule counts, and the vibration has two degrees of freedom, E=k_BT, and this has to be equal to the dissociation energy.

ehild

I have tried finding out frequency using the method you suggested and the value which I end up getting is in the range of 100Khz by using F = (1/2) * 2 \pi √k/m but if I use the quantum definition of energy i.e E= hf , I end up with a value in the range of 10^14 Hz which's sensible.

P.S: I think my first attempt is probably flawed because to calculate mass I have used the following condition: P.E: mgh ( where h = r-r0) to end up with 2.2*10^-11 kg... I have tried it another way and I end up with the same value : f = kx = ma

Can you shed some light on this one ? Thanks

EDIT: Also for my previous answer I get a temperature of about ~12000k , isn't that very unusually high ?

EDIT: It seems I was taking wrong p.e value, I should have used the one in equilibrium. That makes sense but still what could I be doign wrong for the temperature hm...