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, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 3.5 × 10^-3 B) 7.3 × 10^-3 C) 1.4 × 10^-2 D) 2.9 × 10^-3 E) 3.5 × 10^-2

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Quizplus · Accepted Answer

The probability of tunneling through a barrier is given by the formula $ T \approx e^{-2 \kappa a} $, where $ \kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}} $. Substituting the given values, $ \kappa \approx 1.145 \times 10^{10} \, \text{m}^{-1} $ and $ a = 0.40 \, \text{nm} = 0.40 \times 10^{-9} \, \text{m} $, we find $ T \approx e^{-2 \times 1.145 \times 10^{10} \times 0.40 \times 10^{-9}} \approx 3.5 \times 10^{-3} $.

Tunneling: an Electron with Kinetic Energy 2