Tunneling: An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_el = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s) A) 1.4 × 10^-2 B) 2.8 × 10^-2 C) 5.5 × 10^-2 D) 1.1 × 10^-2 E) 1.4 × 10^-1

Question

Quizplus · Accepted Answer

The probability of tunneling can be calculated using the formula: T ≈ e^(-2 * k * a), where k = sqrt(2 * m * (V - E)) / ħ. Substituting the given values, we find that the probability is approximately 1.4 × 10^-2.

Tunneling: an Electron with Kinetic Energy 2