Question

(a) Given that γdrop = 70.30 (mN/m), calculate the radius for a droplet assuming that it is at the Rayleigh limit and contains 1.5 x 104 charges. (b) Now assume that the charges are uniformly distributed on the surface of the droplet and calculate the distance between them. (c) Is this distance consistent with the direct production of a doubly charged small molecule (such as 1,6 diaminohexane) by ion evaporation? Explain why or why not.

Answer 1

A) As we know Rayleigh limit is the limit of maximum charge(Q_ray ) hold by a droplet of radius 'r' . So by the equation Qray=8π√(γε0r³) we can find radius. In equation ε0 =8.854187817...×10−12 F⋅m−1 and called permittivity of free space

Now substituting the value of γ is given as 70.30mN/m,Qray=1.5×10⁴ and ε0 in equation

Therefore 1.5×10⁴=8π√(70.30×8.8×10^-12×r³)

Now taking square both sides to remove square root

(1.5×10⁴)²={8π√(70.30×8.8×10^-12×r³)}²

225×10⁶=631.01×6.186×10^-10×r³

Therefore r³=225×10⁶÷(631.01×6.186×10^-10)

r³=5.76×10¹⁴m.