Question

A beam of electrons moving in the positive \(x\) -direction encounters a potential barrier that is \(2.51 \mathrm{eV}\) high and \(1.00 \mathrm{nm}\) wide. Each electron has a kinetic energy of \(2.50 \mathrm{eV},\) and the electrons arrive at the barrier at a rate of 1000 electrons/s (1000. electrons every second). What is the rate \(\mathrm{I}_{\mathrm{T}}\) in electrons/s at which electrons pass through the barrier, on average? What is the rate \(\mathrm{I}_{\mathrm{R}}\) in electrons/s at which electrons reflect back from the barrier, on average? Determine and compare the wavelengths of the electrons before and after they pass through the barrier.

Short answer

Answer: The rate of electrons passing through the barrier is 196 electrons/s, while the rate of electrons reflecting back is 804 electrons/s. The wavelengths before and after passing through the barrier are approximately the same, which is about 2.6774 x 10^{-11} m.

Step-by-step solution

Find the transmission probability

We can use the formula for transmission probability, \(T\), for a potential barrier: $$ T = \frac{1}{1 + \frac{V_0^2 \sin^2(k_2 a)}{4 E (V_0 - E)}} $$ where: \(V_0\) - potential barrier height \(E\) - electron's kinetic energy \(k_2\) - the wave number inside the barrier \(a\) - barrier width First, we need to convert eV to Joules: 1 eV = 1.60218 × 10^{-19} J \(E = 2.50\) eV = \(4.00545\times 10^{-19}\) J \(V_0 = 2.51\) eV = \(4.01987\times 10^{-19}\) J \(a = 1.00\) nm = \(10^{-9}\) m Now, let's find \(k_2\). We can use the formula: $$ k_2 = \sqrt{ \frac{2m_e (V_0 - E)}{h^2} } $$ where \(m_e\) is the electron mass and \(h\) is the reduced Planck constant: \(m_e = 9.10938 × 10^{-31}\) kg \(h = 1.05457 × 10^{-34}\) Js Plugging the values, we get: $$ k_2 = \sqrt{ \frac{2 \times 9.10938 × 10^{-31} \times (4.01987\times 10^{-19} - 4.00545\times 10^{-19})}{(1.05457 × 10^{-34})^2} } $$ $$ k_2 = 1.22990\times 10^{10} \mathrm{m}^{-1} $$ Now we can calculate the transmission probability, \(T\): $$ T = \frac{1}{1 + \frac{(4.01987\times 10^{-19})^2 \sin^2(1.22990\times 10^{10} \times 10^{-9})}{4 (4.00545\times 10^{-19}) (4.01987\times 10^{-19} - 4.00545\times 10^{-19})}} $$ $$ T \approx 0.196 $$

Determine the rates of electrons passing and reflecting

In order to find the rate of electrons transmitted through the barrier (\(I_T\)) and the reflected rate (\(I_R\)), we use the following formula: \(I_T = I_0 \times T\) \(I_R = I_0 - I_T\) where \(I_0\) is the incoming electron rate and \(T\) is the transmission probability: \(I_0 = 1000\) electrons/s \(T = 0.196\) Now we calculate \(I_T\) and \(I_R\): $$ I_T = 1000 \times 0.196 = 196 \mathrm{\ electrons/s} $$ $$ I_R = 1000 - 196 = 804 \mathrm{\ electrons/s} $$

Determine and compare the wavelengths

We can use the de Broglie wavelength formula to find the wavelengths of the electrons before and after the barrier: $$ \lambda = \frac{h}{p} $$ where \(p\) is the momentum of the electron. We can find \(p\) using the formula: $$ p = \sqrt{2 m_e E} $$ Before the barrier: $$ p = \sqrt{2 \times 9.10938 × 10^{-31} \times 4.00545\times 10^{-19}} = 3.9374\times 10^{-24} \mathrm{kg.m/s} $$ After the barrier (assuming energy is conserved in this case): Solution to this exercise (continue): $$ p = \sqrt{2 \times 9.10938 × 10^{-31} \times 4.00545\times 10^{-19}} = 3.9374\times 10^{-24} \mathrm{kg.m/s} $$ Now we find \(\lambda\): Before the barrier: $$ \lambda = \frac{1.05457 × 10^{-34}}{3.9374\times 10^{-24}} \approx 2.6774\times 10^{-11} \mathrm{m} $$ After the barrier (assuming energy is conserved): $$ \lambda = \frac{1.05457 × 10^{-34}}{3.9374\times 10^{-24}} \approx 2.6774\times 10^{-11} \mathrm{m} $$ As we can see, the wavelengths of the electrons before and after they pass through the barrier are approximately the same.

Key concepts

Potential Barrier

The concept of a potential barrier is a cornerstone in quantum mechanics, representing an energy threshold that particles must overcome to continue their motion. In classical physics, if a particle does not possess enough kinetic energy to surmount this barrier, it will be reflected. However, quantum mechanics introduces a peculiar phenomenon where, under the right conditions, particles can 'tunnel' through the barrier even when their kinetic energy is lower than the barrier's potential energy.

This tunneling is due to the wave-like properties of quantum particles. When we refer to a potential barrier in quantum scenarios, we are often discussing a region of space where the potential energy is higher than the energies of the particles that encounter it. In the case of our exercise, the electrons have a kinetic energy slightly less than the potential barrier, yet some are able to pass through it due to quantum tunneling.

Wave-Particle Duality

Wave-particle duality is one of the principal features of quantum physics. It states that particles such as electrons exhibit both wave-like and particle-like properties. The famous double-slit experiment demonstrates this duality, wherein particles create an interference pattern (a property of waves) when not observed, but appear to travel in distinct paths (as particles) when measured.

The de Broglie hypothesis posits that all matter has a wave associated with it, characterized by the de Broglie wavelength. When dealing with phenomena like quantum tunneling, we consider the wave-like aspect of electrons, which allows us to calculate the probability of them tunneling through a potential barrier. This dual nature of matter is crucial to explaining and predicting outcomes in quantum mechanics.

Transmission Probability

Transmission probability is a measure of how likely a particle is to tunnel through a potential barrier. In quantum mechanics, this is not a binary outcome but a probability, dictated by the wave function of the particle and the properties of the barrier, such as its width and height relative to the particle's energy.

The mathematical expression for the transmission probability involves complex factors such as the incident wave's energy and the characteristics of the barrier. The exercise given demonstrates the calculation of the transmission probability for a beam of electrons encountering a barrier. Here, despite their energy being ever so slightly less than that of the barrier, there exists a non-zero probability (approximately 0.196 in the given example) that the electrons will tunnel through to the other side. In simpler terms, out of 1000 attempts, approximately 196 electrons are expected to make it through the barrier each second.

de Broglie Wavelength

The de Broglie wavelength represents the wave aspect of matter, assigned to particles based on their momentum. According to Louis de Broglie's hypothesis, the wavelength (\f\(\f\text{ambda}\f\)) of a particle is inversely proportional to its momentum (p), formulated as \f\(\f\text{ambda} = \f\f\text{h}/\f\f\text{p}\f\), with \f\f\text{h}\f$ being the reduced Planck constant. This concept allows us to visualize particles like electrons as waves when they interact with potential barriers or other quantum phenomena.

In the context of the exercise, we see that the calculated de Broglie wavelengths before and after the barrier are the same, suggesting that, should the electrons tunnel through, their wavelengths (and thus their momentum, assuming energy conservation) remain constant. This is a key aspect of understanding tunneling, as it ties into the conservation of energy and the probabilistic nature of quantum mechanics.