The potential energy (U) of a diatomic molecule is a function dependent on r (inter atomic distance) as U=αr10−βr5−3 Where, α and β are positive constants. The equilibrium distance between two atoms will be 2αβab, where a=________.

U=ar10−βr5−3At equilibrium F=0⇒F=−dUdr=−−10αr11+5βr6−0⇒0=10αr11−5βr6⇒10αr11=5βr6⇒α=β2r5⇒2αβ=r5⇒r=2αβ1/5 So, a=1 and b=5

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The potential energy (U) of a diatomic molecule is a function dependent on r (inter atomic distance) as
$U = \frac{α}{r^{10}} - \frac{β}{r^{5}} - 3$ Where, $α$ and $β$ are positive constants. The equilibrium distance between two atoms will be ${(\frac{2 α}{β})}^{\frac{a}{b}},$ where $a =________$ .

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Detailed Solution

$U = \frac{a}{r^{10}} - \frac{β}{r^{5}} - 3$

$At equilibrium F=0$

$\begin{array}{l} \Rightarrow F = \frac{- d U}{d r} = - (- \frac{10 α}{r^{11}} + \frac{5 β}{r^{6}} - 0) \\ \Rightarrow 0 = \frac{10 α}{r^{11}} - \frac{5 β}{r^{6}} \\ \Rightarrow \frac{10 α}{r^{11}} = \frac{5 β}{r^{6}} \\ \Rightarrow α = \frac{β}{2} r^{5} \\ \Rightarrow \frac{2 α}{β} = r^{5} \\ \Rightarrow r = {(\frac{2 α}{β})}^{1 / 5} \\ So, a = 1 and b = 5 \end{array}$