Verified Solution • 2025–26 Session moderate

QJEEM20Ph6400: JEE Main 2020 Physics - Class 11 JEEM20Ph6400 Solution: If the potential energy between two molecules is given by U = , …

If the potential energy between two molecules is given by
U = −Ar6+Br12- {A \over {{r^6}}} + {B \over {{r^{12}}}} ,
then at equilibrium, separation between molecules, and the potential energy are :

Multiple Choice Options

  • A (2BA)1/6{\left( {{{2B} \over A}} \right)^{1/6}} , −A24B- {{{A^2}} \over {4B}}
  • B (2BA)1/6,−A22B{\left( {{{2B} \over A}} \right)^{1/6}}, - {{{A^2}} \over {2B}}
  • C (BA)1/6,0{\left( {{B \over A}} \right)^{1/6}},0
  • D (B2A)1/6,−A22B{\left( {{B \over {2A}}} \right)^{1/6}}, - {{{A^2}} \over {2B}}

Step-by-Step Verified Solution

U = −Ar6+Br12- {A \over {{r^6}}} + {B \over {{r^{12}}}}

F = - dUdr{{dU} \over {dr}}

= – (A(–6r–7 )) + B(–12r–13)

for equilibrium, F = 0

∴\therefore 0 = 6Ar7−12Br13{{6A} \over {{r^7}}} - {{12B} \over {{r^{13}}}}

⇒\Rightarrow 6A12B=1r6{{6A} \over {12B}} = {1 \over {{r^6}}}

⇒\Rightarrow r = (2BA)16{\left( {{{2B} \over A}} \right)^{{1 \over 6}}}

∴\therefore U = −A2BA+B(2BA)2- {A \over {{{2B} \over A}}} + {B \over {{{\left( {{{2B} \over A}} \right)}^2}}}

= −A22B+A24B- {{{A^2}} \over {2B}} + {{{A^2}} \over {4B}}

= −A24B- {{{A^2}} \over {4B}}

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