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,h = 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 equation:$$P = e^{-2\kappa L}$$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:$$\kappa = \sqrt{2m(V-E)/\hbar^2}$$where $m$ is the mass of the particle, $V$ 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:$$\kappa = \sqrt{2(1.67 	imes 10^{-27} 	ext{ kg})(6.8 	imes 10^{-15} 	ext{ m})(1.0 	imes 1.60 	imes 10^{-19} 	ext{ J})/(1.055 	imes 10^{-34} 	ext{ J s})^2} = 1.05 	imes 10^{10} 	ext{ m}^{-1}$$So, the probability of tunneling is:$$P = e^{-2(1.05 	imes 10^{10} 	ext{ m}^{-1})(6.8 	imes 10^{-15} 	ext{ m})} = 0.051$$

The Energy of a Proton Is 1