An electron m_e = 9.11 × 10^-31 kg) traveling with a velocity of 1.1 × 10⁶ m/s has a de Broglie wavelength of approximately A) 0.66 nm. B) 0.78 nm. C) 0.89 nm. D) 0.97 nm. E) 1.2 nm.

Question

Quizplus · Accepted Answer

The de Broglie wavelength is given by the equation λ = h/p, where h is Planck's constant and p is the momentum of the particle. The momentum of the electron is given by p = mv, where m is the mass of the electron and v is its velocity. Substituting these values into the equation for the de Broglie wavelength, we get:λ = h/p = h/(mv) = (6.626 × 10^-34 J s)/(9.11 × 10^-31 kg × 1.1 × 10^6 m/s) = 0.66 nmTherefore, the de Broglie wavelength of the electron is approximately 0.66 nm.

An Electron Me = 9

An Electron M_e = 9