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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2022
  • Medium
  • MCQ

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

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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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