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 = 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 can be calculated using the formula for the transmission coefficient T, which is approximately T ≈ e^(-2 * k * a), where k = sqrt((2 * m * (V - E)) / ħ^2), m is the electron mass, V is the barrier height, E is the electron energy, a is the barrier width, and ħ is the reduced Planck's constant. Substituting the given values, the probability is approximately 3.5 × 10^-3.

An Electron with Kinetic Energy 2