In a technique used for surface analysis called Auger electron spectroscopy (AES), electrons are accelerated toward a metal surface. These electrons cause the emissions of secondary electrons—called Auger electrons—from the metal surface. The kinetic energy of the Auger electrons depends on the composition of the surface. The presence of oxygen atoms on the surface results in Auger electrons with a kinetic energy of approximately 506 eV. What is the de Broglie wavelength of one of these electrons? [KE = 1/2mv^2; 1 electron volt (eV) = 1.602 * 10^(-19) J]

Verified step by step guidance

1. Convert the kinetic energy from electron volts (eV) to joules (J) using the conversion factor: 1 eV = 1.602 \times 10^{-19} J.
2. Use the kinetic energy formula KE = \frac{1}{2}mv^2 to solve for the velocity (v) of the electron. Rearrange the formula to v = \sqrt{\frac{2 \times KE}{m}}, where m is the mass of an electron (9.109 \times 10^{-31} kg).
3. Substitute the kinetic energy in joules and the mass of the electron into the velocity equation to find the velocity.
4. Use the de Broglie wavelength formula \lambda = \frac{h}{mv}, where h is Planck's constant (6.626 \times 10^{-34} m^2 kg/s).
5. Substitute the values for Planck's constant, the mass of the electron, and the velocity into the de Broglie wavelength formula to find the wavelength.