MCQ
The potential energy $(U)$ of a diatomic molecule is a function dependent on $r$ (interatomic distance) as

$U =\frac{\alpha}{ r ^{10}}-\frac{\beta}{ r ^{5}}-3$

where, $\alpha$ and $\beta$ are positive constants. The equilibrium distance between two atoms will be $\left(\frac{2 \alpha}{\beta}\right)^{\frac{a}{b}},$ where a $=$ ..........

  • A
    $2$
  • $1$
  • C
    $3$
  • D
    $0$

Answer

Correct option: B.
$1$
b
For equilibrium

$\frac{ dU }{ dr }=0$

$\frac{-10 \alpha}{r^{11}}+\frac{5 \beta}{r^{6}}=0$

$\frac{5 \beta}{r^{6}}=\frac{10 \alpha}{r^{11}}$

$r ^{3}=\frac{2 \alpha}{\beta}$

$r=\left(\frac{2 \alpha}{\beta}\right)^{\frac{1}{3}}$

$a=1$

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