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
Menu
Grade: 11
        
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 ?
8 months ago

Answers : (1)

Arun
21637 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
8 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details