## College Physics (7th Edition)

$1.5×10^{4}$ V
Recall the relation λ= $\frac{1.227}{\sqrt V}nm$ where λ is the wavelength and V is the magnitude of accelerating potential. Thus we get, V=$(\frac{1.227}{ λ(nm)})^{2}$= $(\frac{1.227}{0.010})^{2}$= $1.5×10^{4}$ volts