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(II) A proton and a helium nucleus approach a 25-MeV potential energy barrier. If each has a kinetic energy of \(5.0 \mathrm{MeV}\) what is the probability of each to tunnel through the barrier, assuming it is rectangular and \(3.6 \mathrm{fm}\) thick?

Short Answer

Expert verified
The tunneling probability is approximately 0.0206 for a proton and 0.00043 for a helium nucleus.

Step by step solution

01

Understand the Problem

We are asked to find the probability that a proton and a helium nucleus will tunnel through a potential energy barrier. The given data includes a potential energy barrier of 25 MeV, kinetic energy of 5 MeV for each particle, and a barrier thickness of 3.6 fm.
02

Identify the Relevant Formula

The probability of tunneling through a potential barrier can be calculated using the formula for the transmission coefficient for a rectangular barrier: \[ T = e^{-2\gamma a} \] where \( \gamma = \frac{\sqrt{2m(U-E)}}{\hbar} \), \( a \) is the barrier thickness, \( U \) is the potential energy of the barrier, \( E \) is the kinetic energy of the particle, and \( m \) is the mass of the particle.
03

Calculate Gamma (\(\gamma\)) for Proton

For a proton, the mass \( m_p = 938.3 \text{ MeV}/c^2 \). The barrier height \( U = 25 \text{ MeV} \) and the kinetic energy \( E = 5 \text{ MeV} \).\[ \gamma = \frac{\sqrt{2 \times 938.3 \times (25-5)}}{197.3} \approx 0.539 \text{ fm}^{-1} \]
04

Calculate Tunneling Probability for Proton

Using the calculated \( \gamma \) for the proton:\[ T_p = e^{-2 \times 0.539 \times 3.6} \approx e^{-3.8784} \approx 0.0206 \]
05

Calculate Gamma (\(\gamma\)) for Helium Nucleus

For a helium nucleus (mass \( m_{He} = 4 \times 938.3 \text{ MeV}/c^2 \)), \[ \gamma = \frac{\sqrt{2 \times 4 \times 938.3 \times (25-5)}}{197.3} \approx 1.079 \text{ fm}^{-1} \]
06

Calculate Tunneling Probability for Helium Nucleus

Using the calculated \( \gamma \) for the helium nucleus:\[ T_{He} = e^{-2 \times 1.079 \times 3.6} \approx e^{-7.7608} \approx 0.00043 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transmission Coefficient
Understanding the transmission coefficient is key when discussing quantum tunneling. It represents the probability that a particle will tunnel through a potential energy barrier rather than being reflected back. One can calculate it using the formula:
  • \( T = e^{-2\gamma a} \)
Here, \( T \) stands for the transmission coefficient, \( \gamma \) is the decay constant that depends on the mass of the particle, its kinetic energy, and the height of the barrier, and \( a \) is the thickness of the barrier. According to quantum mechanics, even if a particle does not have enough energy to climb over a barrier, it still has a probability—represented by the transmission coefficient—of appearing on the other side through quantum tunneling.
In the exercise, the probability turns out to be significantly higher for the proton compared to the helium nucleus due to its smaller mass. This indicates that lighter particles have a higher chance of tunneling through barriers compared to heavier ones.
Kinetic Energy
Kinetic energy is all about motion. It's the energy that a particle possesses due to its movement. In quantum tunneling, a particle's kinetic energy \( E \) influences its ability to penetrate a potential energy barrier. If the kinetic energy is lower than the potential energy of the barrier, classical physics suggests the particle cannot pass through. However, quantum tunneling challenges this notion allowing even such particles a non-zero probability of crossing the barrier. In our example, both the proton and helium nucleus have kinetic energies of 5 MeV, which is significantly less than the 25 MeV barrier they encounter. Their kinetic energy plays a vital role in calculating the decay constant \( \gamma \). As the kinetic energy increases, the probability of tunneling also increases, illustrating that a balance of mass, energy, and barrier characteristics dictate tunneling likelihood.
Potential Energy Barrier
A potential energy barrier is essentially an obstacle in the quantum world. It's a region where the potential energy \( U \) is higher than the kinetic energy of particles approaching it. Normal intuition tells us particles cannot surpass this barrier unless their kinetic energy exceeds the energy barrier. However, quantum mechanics offers the peculiar phenomenon of tunneling.In the setup with a 25 MeV potential energy barrier, we explore this concept. Despite being 5 times the kinetic energy of the proton and helium nucleus (5 MeV), both particles still have a chance of tunneling through. Barrier thickness \( a \) is also crucial, as it directly influences the probability; a larger thickness decreases the odds of tunneling. Analyzing such barriers highlights not just the fascinating contrasts between classical and quantum physics but also the influence of parameters like particle mass and energy on quantum behavior.

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Most popular questions from this chapter

(II) An electron with \(180 \mathrm{eV}\) of kinetic energy in free space passes over a finite potential well \(56 \mathrm{eV}\) deep that stretches from \(x=0\) to \(x=0.50 \mathrm{nm}\). What is the electron's wavelength (a) in free space, \((b)\) when over the well? ( \(c\) ) Draw a diagram showing the potential energy and total energy as a function of \(x\), and on the diagram sketch a possible wave function.

Estimate the lowest possible energy of a neutron contained in a typical nucleus of radius \(1.2 \times 10^{-15} \mathrm{~m}\). [Hint: A particle can have an energy at least as large as its uncertainty.

(II) An electron approaches a potential barrier \(18 \mathrm{eV}\) high and \(0.55 \mathrm{nm}\) wide. If the electron has a \(1.0 \%\) probability of tunneling through the barrier, what is the electron's energy?

(III) Consider a particle of mass \(m\) and energy \(E\) traveling to the right where it encounters a narrow potential barrier of height \(U_{0}\) and width \(\ell\) as shown in Fig. \(38-21 .\) It can be shown that: (i) for \(EU_{0},\) the transmission probability is $$ T=\left[1+\frac{\sin ^{2}\left(G^{\prime} \ell\right)}{4\left(E / U_{0}\right)\left(E / U_{0}-1\right)}\right]^{-1} $$ where $$ G^{\prime}=\sqrt{\frac{2 m\left(E-U_{0}\right)}{\hbar^{2}}} $$ and \(R=1-T .\) Consider that the particle is an electron and it is incident on a rectangular barrier of height \(U_{0}=10 \mathrm{eV}\) and width \(\ell=1.0 \times 10^{-10} \mathrm{~m} .\) ( \(a\) ) Calculate \(T\) and \(R\) for the electron from \(E / U_{0}=0\) to \(10,\) in steps of \(0.1 .\) Make a single graph showing the two curves of \(T\) and \(R\) as a function of \(E / U_{0}\). (b) From the graph determine the energies \(\left(E / U_{0}\right)\) at which the electron will have transmission probabilities of \(10 \%\) $$ 20 \%, 50 \%, \text { and } 80 \% $$

(II) Consider a single oxygen molecule confined in a onedimensional rigid box of width \(4.0 \mathrm{~mm}\). \((a)\) Treating this as a particle in a rigid box, determine the ground-state energy. ( \(b\) ) If the molecule has an energy equal to the onedimensional average thermal energy \(\frac{1}{2} k T\) at \(T=300 \mathrm{~K},\) what is the quantum number \(n ?(c)\) What is the energy difference between the \(n\) th state and the next higher state?

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