$U =\frac{\alpha}{ r ^{10}}-\frac{\beta}{ r ^{5}}-3$
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 $=$ ..........
$\frac{ dU }{ dr }=0$
$\frac{-10 \alpha}{r^{11}}+\frac{5 \beta}{r^{6}}=0$
$\frac{5 \beta}{r^{6}}=\frac{10 \alpha}{r^{11}}$
$r ^{3}=\frac{2 \alpha}{\beta}$
$r=\left(\frac{2 \alpha}{\beta}\right)^{\frac{1}{3}}$
$a=1$
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| Column $-I$ | Column $-II$ |
|---|---|
| $(i)$ Velocity of image | $(a) \;2\;m/s$ |
| $(ii)$ Velocity of image with respect to mirror | $(b)\; 20\; m/s$ |
| $(iii)$ Velocity of image with respect to object | $(c) \;11 \;m/s$ |
| $(iv)$ Velocity of image if mirror is stoped | $(d)\; 22\; m/s$ |
(Mass of electron $=9.1 \times 10^{-31} \,{kg}$ )