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 October 17th, 2016, 04:52 AM #1 Newbie   Joined: Oct 2016 From: United Kingdom Posts: 13 Thanks: 0 Stationary Points The potential energy of two molecules separated by a distance rr is given by U=U0((a/r)12−2(a/r)6)U=U0((a/r)12−2(a/r)6) where U0U0 and aa are positive constants. The equilibrium separation of the two molecules occurs when the potential energy is a minimum; find the equilibrium separation and give the value of the potential energy at this separation if U0=6.62×10−21JU0=6.62×10−21J. i dont know how to start this question: so any help is greatly appreciated! October 17th, 2016, 06:21 AM   #2
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Quote:
 Originally Posted by Tom2000 The potential energy of two molecules separated by a distance rr is given by U=U0((a/r)12−2(a/r)6)U=U0((a/r)12−2(a/r)6) where U0U0 and aa are positive constants. The equilibrium separation of the two molecules occurs when the potential energy is a minimum; find the equilibrium separation and give the value of the potential energy at this separation if U0=6.62×10−21JU0=6.62×10−21J.

$U=U_0\bigg[\left(\dfrac{a}{r}\right)^{12}-2\left(\dfrac{a}{r}\right)^6 \bigg]$

In any case, start by finding where $\dfrac{dU}{dr}=0$ to locate possible extrema. October 17th, 2016, 06:47 AM #3 Newbie   Joined: Oct 2016 From: United Kingdom Posts: 13 Thanks: 0 So where would i go from there? October 17th, 2016, 07:07 AM   #4
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Quote:
 Originally Posted by Tom2000 So where would i go from there?
Find the radius at the max/min and resubstitute it into the formula. October 17th, 2016, 07:08 AM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 once you find the value(s) of $r$ that make $\dfrac{dU}{dr}=0$, you can use the first derivative test or the second derivative test to classify the extrema. October 17th, 2016, 07:18 AM #6 Newbie   Joined: Oct 2016 From: United Kingdom Posts: 13 Thanks: 0 How would i find r (really sorry im not very good at maths) October 17th, 2016, 07:25 AM   #7
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Quote:
 Originally Posted by Tom2000 How would i find r (really sorry im not very good at maths)
Hmmm...

Have a go at the following question

Let:
$\displaystyle y = 3x^{12} + 5x^6$

Can you:
a) find $\displaystyle \frac{dy}{dx}$; then
b) find the x-value where $\displaystyle \frac{dy}{dx} = 0$?

If you find this question tricky, we can help you with this  October 17th, 2016, 07:27 AM   #8
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Quote:
 Originally Posted by Tom2000 How would i find r (really sorry im not very good at maths)
what did you get for $\dfrac{dU}{dr}$ ? October 17th, 2016, 07:39 AM   #9
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 Originally Posted by Benit13 Hmmm... Have a go at the following question Let: $\displaystyle y = 3x^{12} + 5x^6$ Can you: a) find $\displaystyle \frac{dy}{dx}$; then b) find the x-value where $\displaystyle \frac{dy}{dx} = 0$? If you find this question tricky, we can help you with this so dy/dx = 36x^11+30x^5

how would i do the next part? October 17th, 2016, 07:41 AM   #10
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Quote:
 Originally Posted by Tom2000 so dy/dx = 36x^11+30x^5 how would i do the next part?

Good start!

The maximum or minimum of that function occurs when the gradient (dy/dx) is zero:

dy/dx = 0

so try solving the equation

36x^11+30x^5 = 0

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