JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
Potential energy as a function of \(r\) is given by \(U=\frac{ A }{ r ^{10}}-\frac{ B }{ r ^{5}}\), where \(r\) is the interatomic distance, \(A \) and \(B\) are positive constants. The equilibrium distance between the two atoms will be
- A \(\left(\frac{ A }{ B }\right)^{\frac{1}{5}}\)
- B \(\left(\frac{B}{A}\right)^{\frac{1}{5}}\)
- C \(\left(\frac{2 A }{ B }\right)^{\frac{1}{5}}\)
- D \(\left(\frac{B}{2 A}\right)^{\frac{1}{5}}\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{2 A }{ B }\right)^{\frac{1}{5}}\)
Step-by-step Solution
Detailed explanation
\(\frac{-10\, A }{ r ^{11}}+\frac{5\,B }{ r ^{6}}=0\) \(r ^{5}=\frac{10\, A }{5 \,B }=\frac{2 \,A }{ B }\)
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