The de Broglie Wavelength ($\lambda$) of an accelerated electron is given by: $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2meV}}$ Using standard values for $h,m,e$, the approximate formula is: $\lambda \approx \sqrt{\frac{150}{V}}\text{ A˚}\approx \frac{12.27}{\sqrt{V}}\text{ A˚}$ Given $V=100\text{ V}$: $\lambda =\frac{12.27}{\sqrt{100}}=\frac{12.27}{10}=1.227\text{ A˚}$ Since $1\text{ nm}=10\text{ A˚}$: $\lambda \approx 0.1227\text{ nm}\approx 0.123\text{ nm}$