The liquid drop needs latent heat to evaporate. This latent heat of vaporization at a constant temperature is supplied by the surface energy.
Let, \( r \) be the radius of a liquid drop, \( T \) be the surface tension of the liquid, then the surface energy associated with this liquid drop is given by,

Surface energy= surface area \( \times \) its surface tension = \( 4\pi{r^2}\cdot{T} \).

The amount of heat required for vaporization of the liquid drop is \( \frac{4}{3}\pi{r^3}L \), where \( L \) is the latent heat of vaporization of the liquid per unit volume.

Therefore, \( 4\pi{r^2}\cdot{T}=\frac{4}{3}\pi{r^3}L \)

or, \( \displaystyle{r=\frac{3T}{L}} \)

So the critical radius of a liquid drop to evaporate is \( \displaystyle{\frac{3T}{L}} \) and all the drops of this critical size or less that this critical size will autometically evaporate.