Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        The potential energy between two atoms in a molecule given by U(x)=a/x12 – b/x6 where a & b are positive constants and x is distance between atoms . The atom is in stable equilibrium when x is equal to what ?`
11 months ago

Arun
22615 Points
```							1)dU/dr = -12a/(r^13) +6b/(r^7) = 0 or 12 a = 6b*(r^6)b or r = {(2)^(1/6)}*(a/b)^(1/6)} at equilibrium. differentiating once again we find that d^2/U/dr^2 > 0 for r = {(2)^(1/6)}*(a/b)^(1/6) Umin = a/{4*(a/b)^2} - b/{2*(a/b)} = b^2/(4a) - b^2/(2a) = -(b^2)/4a So required minimum energy = (b^2)/4a 2)and 3) {(2)^(1/6)}*(a/b)^(1/6)} = 1.13*10^(-10) or (a/b) =(1/2) (1.13)^6 * 10^-60 = 1.041*10^-60 ----------------- 1, and (b^2)/4a = 1.54*10^-18 or b = 4*1.54*(10^-18)*(a/b) ---------------------- 2; substituting from 1 fro a/b, weget] b = 6.16*(10^-18)*1.041*10^-60 = 6.142*10^-78 J m^6 and  a = 6.675*10^-138 j m^12
```
11 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions