Chapter 30 - Quantum Physics - Problems and Conceptual Exercises - Page 1076: 91

$0.58~nm$

We can find the minimum de Broglie wavelength as follows: $\lambda=\frac{h}{\sqrt{2m(hf-W_{\circ})}}$ We plug in the known values to obtain: $\lambda=\frac{6.63\times 10^{-34}}{\sqrt{2(9.11\times 10^{-31})[(6.63\times 10^{-34})(2.11\times 10^{15})-(6.8\times 10^{-19})]}}$ $\lambda=0.57\times 10^{-9}m=0.58~nm$