Chapter 28 - Quantum Mechanics and Atomic Physics - Learning Path Questions and Exercises - Exercises - Page 962: 5

$1.5×10^{4}$ 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