A 3.10-eV electron is incident on a 0.40-nm barrier that is 5.67 eV high.What is the probability that this 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)1.4 × 10^-3 B)9.0 × 10^-4 C)1.0 × 10^-3 D)1.5 × 10^-3 E)1.2 × 10^-3

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Quizplus · Accepted Answer

The probability of tunneling is given by the following equation:$$T = e^{-2\kappa L}$$where $\kappa$ is the wave vector of the electron in the barrier and $L$ is the width of the barrier. The wave vector is given by:$$\kappa = \sqrt{2m(V_0 - E)}/\hbar$$where $m$ is the mass of the electron, $V_0$ is the height of the barrier, and $E$ is the energy of the electron.Plugging in the given values, we get:$$\kappa = \sqrt{2(9.11 	imes 10^{-31} 	ext{ kg})(5.67 	ext{ eV} - 3.10 	ext{ eV})}/(1.055 	imes 10^{-34} 	ext{ J s})}$$$$\kappa = 1.05 	imes 10^{10} 	ext{ m}^{-1}$$The width of the barrier is given as $0.40 	ext{ nm} = 4.00 	imes 10^{-10} 	ext{ m}$.Plugging these values into the equation for the probability of tunneling, we get:$$T = e^{-2(1.05 	imes 10^{10} 	ext{ m}^{-1})(4.00 	imes 10^{-10} 	ext{ m})}$$$$T = 1.4 	imes 10^{-3}$$Therefore, the probability that the electron will tunnel through the barrier is $1.4 	imes 10^{-3}$.

A 310-EV Electron Is Incident on a 0