Answer
$1.02\times 10^{-19}$ $J$
Work Step by Step
The total work needed to stretch a bond by a displacement $x$ is given by
$W=\frac{1}{2}kx^2$
where $k$ is force constant.
The force constant of a $H–H$ bond is given: $k= 510$ $Nm^{−1}$
Now the work needed to stretch the bond by $x=20$ $pm$ is
$W=\frac{1}{2}\times 510\times (20\times 10^{-12})^2$ $J$ $=1.02\times 10^{-19}$ $J$
