Answer
$2.5\times10^{-19}J$
Work Step by Step
Given that the $F=-kx+bx^{3}$ where, F - Force, k - 0.113 nN/nm, $b - 0.00185\space nN/nm^{3}$, x - Stretch of the electron.
This bond isn't a simple $F=-kx$ spring for which we already have a potential - energy formula. Because the bond force varies with stretch, we'll have to integrate. We can use the following equation for stored potential energy.
$\Delta U= -\int_{x_{1}}^{x_{2}}F(x)dx$ ; Let's plug known values into this equation.
$U=-\int_{0}^{x}(-kx+bx^{3})\space dx= \frac{1}{2}kx^{2}-\frac{b}{4}x^{4}\Biggr |_{0}^{x} $
$U=\frac{1}{2}kx^{2}-\frac{b}{4}x^{4}$
$U=\frac{1}{2}\times 0.113\space nN/nm\times (2.14nm)^{2}-\frac{0.00185\space nN/nm^{3}}{4}(2.14nm)^{4}$
$U=0.26\space nNnm\space -0.01\space nNnm = 0.25\times10^{-9} J$
$U=2.5\times10^{-19}J$
