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An electron with initial kinetic energy 5.5 eV encounters a square potential barrier of height 10.0 eV. What is the width of the barrier if the electron has a 0.50\(\%\) probability of tunneling through the barrier?

Short Answer

Expert verified
The width of the barrier is approximately 2.32 x 10^{-10} m.

Step by step solution

01

Determine the difference in energy

The electron's initial kinetic energy is 5.5 eV, and the barrier's height is 10.0 eV. Thus, the energy difference that the electron needs to overcome is given by:\[ V - E = 10.0 \text{ eV} - 5.5 \text{ eV} = 4.5 \text{ eV} \]
02

Use the tunneling probability formula

The tunneling probability for a particle encountering a potential barrier can be estimated using the formula: \[ T \approx e^{-2 \kappa a} \]where \( T \) is the tunneling probability, \( \kappa = \frac{\sqrt{2m(U-E)}}{\hbar} \), \( a \) is the barrier width, and \( U-E = 4.5 \text{ eV} \) as found in Step 1.
03

Calculate \( \kappa \)

First, convert the energy difference from eV to joules using the conversion: \[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]Thus, \[ U-E = 4.5 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} \approx 7.209 \times 10^{-19} \text{ J} \]Now calculate \( \kappa \) using:\[ \kappa = \frac{\sqrt{2m(U-E)}}{\hbar} \]where \( m = 9.109 \times 10^{-31} \text{ kg} \) (mass of an electron) and \( \hbar = 1.055 \times 10^{-34} \text{ Js} \). Substitute:\[ \kappa = \frac{\sqrt{2 \times 9.109 \times 10^{-31} \text{ kg} \times 7.209 \times 10^{-19} \text{ J}}}{1.055 \times 10^{-34} \text{ Js}} \approx 1.14 \times 10^{10} \text{ m}^{-1} \]
04

Solve for barrier width \( a \)

We know that the electron has a 0.50% probability of tunneling, or \( T = 0.005 \). Thus:\[ 0.005 = e^{-2 \times 1.14 \times 10^{10} \times a} \]Take the natural logarithm of both sides:\[ \ln(0.005) = -2 \times 1.14 \times 10^{10} \times a \]Thus:\[ a = \frac{\ln(0.005)}{-2 \times 1.14 \times 10^{10}} \approx \frac{-5.298}{-2.28 \times 10^{10}} \approx 2.32 \times 10^{-10} \text{ m} \]
05

Conclusion

The width of the potential barrier is approximately \( 2.32 \times 10^{-10} \text{ m} \).

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Key Concepts

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

potential barrier
In the world of quantum mechanics, a potential barrier is a region where the potential energy of a particle, like an electron, is higher than the surrounding regions. Imagine you're playing a video game, and your character encounters a magic wall that seems impassable. For the electron, this wall is the potential barrier. Even though classical mechanics suggests that the electron should bounce back if it doesn't have enough energy, quantum mechanics opens up a unique possibility called "tunneling." - A potential barrier has a specific height, measured in energy units like electron volts (eV). - The height of this barrier determines how much energy a particle needs to overcome it. In our example, the barrier is 10 eV tall. - If the electron's energy is less than this height (as is the case here, with the electron having 5.5 eV), it can still "tunnel" through the barrier, thanks to quantum effects. The concept of the potential barrier is essential for understanding why electrons can sometimes seem to perform a magical act, passing right through solid obstacles.
kinetic energy
Kinetic energy is the energy a particle possesses due to its motion. For an electron, this is a measure of its speed and, indirectly, its ability to overcome obstacles like potential barriers. - Think of kinetic energy as the electron's fuel for the journey across barriers. - It's typically measured in electron volts (eV). In this scenario, the electron starts with a kinetic energy of 5.5 eV. Now, why is kinetic energy so important here? It's because the amount of kinetic energy determines if the electron will even try to tackle the barrier. The higher the electron's kinetic energy, the better its chances of tunneling through the barrier, even if that energy isn't enough by classical standards to go over it. But as we'll see, tunneling doesn't care much for classical mechanics!
tunneling probability
Tunneling probability is a fascinating concept that describes the likelihood of a particle like an electron penetrating through a potential barrier even when its kinetic energy is less than the barrier height.- Think of it as the odds of our electron managing to slip through the barrier's gaps. - It's mostly a number between 0 and 1, where 0 means no chance of tunneling and 1 implies certainty.In our scenario, the exercise calculates a tunneling probability of 0.50%, or 0.005. Quantum mechanics provides a formula to approximate this probability: \[ T \approx e^{-2 \kappa a} \]Here, \(T\) is the tunneling probability, \(\kappa\) is a constant related to the barrier properties and the electron, and \(a\) is the barrier width. This formula encapsulates how tunneling probability decreases exponentially with increased barrier width or electron mass.
electron mass
The mass of an electron plays a crucial role in quantum tunneling and in determining the value of \(\kappa\) in the tunneling probability formula. It might seem like just another number, but this tiny mass has big effects when it comes to the tunneling process!- The electron mass is approximately \(9.109 \times 10^{-31} \text{ kg}\).- It's a factor in calculating \(\kappa\), which contains terms that include the mass and energy values.In quantum mechanics, because the electron has such a small mass, its wave-like properties are amplified, making tunneling through barriers a common phenomenon. This is why light particles like electrons can exhibit tunneling more frequently and noticeably than heftier particles. It's the combination of the tiny size and mass that allows electrons to "tunnel" through even seemingly insurmountable barriers.

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Most popular questions from this chapter

Consider a wave function given by \(\psi(x) = A \space sin \space kx\), where \(k = 2\pi/\lambda\) and \(A\) is a real constant. (a) For what values of \(x\) is there the highest probability of finding the particle described by this wave function? Explain. (b) For which values of \(x\) is the probability \(zero\)? Explain.

Consider the wave packet defined by $$\psi(x) = \int ^\infty_0 B(k)cos kx dk$$ Let \(B(k) = e^{-a^2k^2}\). (a) The function \(B(k)\) has its maximum value at \(k\) = 0. Let \(k_h\) be the value of \(k\) at which \(B(k)\) has fallen to half its maximum value, and define the width of \(B(k)\) as \(w_k = k_h\) . In terms of \(\alpha\), what is \(w_k\) ? (b) Use integral tables to evaluate the integral that gives \(\psi(x)\). For what value of \(x\) is \(\psi(x)\) maximum? (c) Define the width of \(\psi(x)\) as \(w_x = x_h\) , where \(x_h\) is the positive value of \(x\) at which \(\psi(x)\) has fallen to half its maximum value. Calculate \(w_x\) in terms of \(\alpha\). (d) The momentum \(p\) is equal to \(hk/2\pi\), so the width of \(B\) in momentum is \(w_p = hw_k /2\pi\). Calculate the product \(w_p w_x\) and compare to the Heisenberg uncertainty principle.

One advantage of the quantum dot is that, compared to many other fluorescent materials, excited states have relatively long lifetimes (10 ns). What does this mean for the spread in the energy of the photons emitted by quantum dots? (a) Quantum dots emit photons of more well-defined energies than do other fluorescent materials. (b) Quantum dots emit photons of less well-defined energies than do other fluorescent materials. (c) The spread in the energy is affected by the size of the dot, not by the lifetime. (d) There is no spread in the energy of the emitted photons, regardless of the lifetime.

Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width 0.360 nm. (b) The electron makes a transition from the \(n\) = 1 to \(n\) = 4 level by absorbing a photon. Calculate the wavelength of this photon.

A harmonic oscillator absorbs a photon of wavelength 6.35 \(\mu\)m when it undergoes a transition from the ground state to the first excited state. What is the ground-state energy, in electron volts, of the oscillator?

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