## University Physics with Modern Physics (14th Edition)

$3.43\times 10^{-10}m$.
Figure 40.15b in the textbook shows values for the bound state energy of a square well in the situation $U_o=6E_{1-IDW}$. We know $E_{1-IDW}=\frac{ h^2}{8mL^2}$. From the figure we see that $E_1=0.625E_{1-IDW}=0.625\frac{ h^2}{8mL^2}$. Solve for the width L, using $E_1=2.00eV=3.20\times10^{-19}J$. $$3.20\times10^{-19}J =0.625\frac{ h^2}{8mL^2}$$ $$L=3.43\times 10^{-10}m$$