Tunneling: The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier. What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg, ħ = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s) A) 0.051 B) 0.048 C) 0.056 D) 0.053

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Quizplus · Accepted Answer

The probability of tunneling is given by the formula:$$P = e^{-2\kappa L}$$where $\kappa$ is the wave number of the particle in the barrier and $L$ is the width of the barrier.The wave number is given by:$$\kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}$$where $m$ is the mass of the particle, $V_0$ is the height of the barrier, $E$ is the energy of the particle, and $\hbar$ is the reduced Planck constant.Plugging in the given values, we get:$$\kappa = \sqrt{\frac{2(1.67 	imes 10^{-27} 	ext{ kg})(1.0 	ext{ MeV})(1.60 	imes 10^{-19} 	ext{ J/eV})}{(1.055 	imes 10^{-34} 	ext{ J s})^2}} = 1.05 	imes 10^{11} 	ext{ m}^{-1}$$The width of the barrier is given as $L = 6.8 	ext{ fm} = 6.8 	imes 10^{-15} 	ext{ m}$.Plugging these values into the formula for the probability of tunneling, we get:$$P = e^{-2(1.05 	imes 10^{11} 	ext{ m}^{-1})(6.8 	imes 10^{-15} 	ext{ m})} = 0.051$$Therefore, the probability that the proton will tunnel through the barrier is 0.051.

Tunneling: the Energy of a Proton Is 1