What is the de brogile wavelength of an electron in Armstrong at 100 ev?

The de Broglie wavelength (\lambda) of a particle can be calculated using the formula (\lambda = \frac{h}{p}), where (h) is Planck's constant and (p) is the momentum. For an electron with kinetic energy of 100 eV, its momentum can be derived from the relation (p = \sqrt{2mK}), where (m) is the electron mass and (K) is the kinetic energy in joules. Converting 100 eV to joules (1 eV = (1.6 \times 10^{-19}) J), and using the values for (h) and (m), the de Broglie wavelength is approximately 12.3 picometers, or 0.123 angstroms.