# Homework Help: Interatomic potential and binding energy system of equations

1. Mar 26, 2014

### KiNGGeexD

I have attached a photograph of my problem to save time on typing the question

I have done the first part to obtain the expression for equilibrium bond length and in the second part I solved that equation for A and substituted it into my binding energy equation and obtained an answer with the order of magnitude to the -60 so when I square it in my calculator I get 0, I'm assuming this is incorrect somehow although I don't see what I have done wrong

Thanks for any help in advanced, all necessary equations are present in the problem

2. Mar 26, 2014

### BvU

Broke my screen trying to read your picture without breaking my neck.
Please use the template. Read the guidelines to find out why saving your time typing is no good. Let alone breaking helpers' screens and/or necks!

Show your calculations. Upright. Then we'll check

3. Mar 26, 2014

### KiNGGeexD

Sorry I will post another photograph and will take a photo also of my calculations just now

4. Mar 26, 2014

### KiNGGeexD

5. Mar 26, 2014

### KiNGGeexD

This is what I have got! But I keep getting A as 0

6. Mar 26, 2014

### KiNGGeexD

Regardless of how I do it my numbers are far to small

7. Mar 27, 2014

### BvU

If your calculator can't handle these small numbers, you can always leave out the powers of ten, calculate, and then add the powers of ten again -- by hand ! Not so difficult, actually.

Work a little more neatly. $B^2/{4A}$ is not 0.0031, so B is even smaller than you thought. Let alone A !

Still, the answer is correct. Remember that 1/r0^12 is huge !

8. Mar 27, 2014

### KiNGGeexD

Ok I will give it a bash

9. Mar 27, 2014

### KiNGGeexD

Ok so I recalculated for A using B as just 2.63 and leaving out the negative power, I obtained A to be 557.81

Now my only issue is powers, when I use these constants in the first equation though just as they are they work, because the ratios are the same but they don't in the second equation

10. Mar 27, 2014

### KiNGGeexD

Sorry excuse my idiocy it does also work in the second equation :(

So my only issues is powers, one is to the negative 60 so if the ratio with the two numbers work then surely I have 557.81 to the negative 60 also,

Or
5.5781 to the negative 62?

11. Mar 27, 2014

### KiNGGeexD

To the negative 58 rather, as it is larger haha

12. Mar 27, 2014

### KiNGGeexD

Ok I redone all of my maths and wrote it out properly and got A= 5.576 to the negative 118

When I put that back into my expression and used the laws of indices to make it simpler I got the correct answer which makes sense considering my powers which where then squared etc

I do have a question regarding these constants? They are tiny... I mean an atom is what tens of femto metres so something to the negative 118 is over 100 orders of magnitude smaller

13. Mar 27, 2014

### BvU

Six posts! Which one do you want me to read ?

In the mean time, maybe you can read post 7
It has some clue as to why A needs to be pretty small.

14. Mar 27, 2014

### KiNGGeexD

I know sorry I post and change my mind gaha but the post with A to the negative 118 is the right answers I'm sure I tried it in the equation

15. Mar 27, 2014

### dauto

It's much easier to use nm as your unit for distances instead of meters that way you avoid having to deal with the powers of 10 all together.

16. Mar 27, 2014

### dauto

Look at the units. A and B are not sizes. Their units are eV nm6 and eV nm12. You really should use nm instead of meters for that problem.

17. Mar 27, 2014

### KiNGGeexD

I know they aren't sizes just extremely small I was just trying to get a sense of proportion as if they were sizes :)

18. Mar 27, 2014

### dauto

But that doesn't work. The numbers in themselves are completely meaningless without the units and with the units it can only be compared to other quantities that have the same units. Mentioning the size of an atom without the units will not help you understand anything.

19. Mar 27, 2014

### KiNGGeexD

Ok sorry haha!!! Are the constants ok though in terms of validity of the problem

20. Mar 27, 2014

### BvU

Good thing to try and assimilate some of the stuff just worked out. Physicist in the making! Start by googling Lennard-Jones potential.

Just so you know we're all human: I looked there first too, so I could assist without making too much of a fool of myself again