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An electron is bound in a square well of width \(1.50 \mathrm{nm}\) and 40.25 depth \(U_{0}=6 E_{1-\mathrm{IDW}}\). If the electron is initially in the ground level and absorbs a photon, what maximum wavelength can the photon have and still liberate the electron from the well?

Short Answer

Expert verified
By following the steps outlined here, the maximum wavelength λ of the photon that would liberate the electron from the well may be calculated. The exact value would require plugging in the calculated values of \(E_{1}\) and \(U_{0}\) from the early steps and solving the final equation.

Step by step solution

01

Understanding the Energy Levels of a Quantum Mechanical Square Well

A quantum mechanical square well is a potential well that has definite energy levels, which can be calculated by the equation \(E_{n} =\frac{n^{2}h^{2}}{8mL^{2}}\), where, \(E_{n}\) is the energy of the state, \(n\) is the quantum number of the state (typically starting at 1 for the ground state), \(h\) is the Planck constant divided by \(2Ï€\), \(m\) is the electron mass and \(L\) is the width of the well. In this case, the electron is in the ground state i.e., \(n = 1\) and \(L = 1.50 nm\). and the energy is given as \(6E_{1-IDW}\). So, we can calculate \(E_{1}\).
02

Calculate the Ground State Energy

To calculate the ground state energy \(E_{1}\), we use the equation presented in step one, plug in the known values and solve for \(E_{1}\). Substituting \(n = 1\), \(L = 1.50 \times 10^{-9} m\) (since 1 nm = \(10^{-9} m\)),\(h = 6.62607004 \times 10^{-34} m^{2} kg / s\), \(m = 9.10938356 \times 10^{-31} kg\) (mass of an electron), we get\[E_{1} = \frac{6.62607004^{2} × 10^{-34} × 10^{-34}}{8 × 9.10938356 × 10^{-31} × (1.50 × 10^{-9})^{2}}\]
03

Calculate the Well Depth

The depth of well \(\(U_{0}\)\) is given by \(U_{0}=6 E_{1-\mathrm{IDW}}\). Therefore, we plug the value of \(E_{1}\) calculated in the previous step into this formula to find \(U_{0}\).
04

Calculate the Maximum Wavelength

To liberate the electron from the well, the energy of the photon (\(E = h c / λ\)) must be more than the well depth (\(U_{0}\)). So we solve \(h c / λ = U_{0}\) for λ to find the maximum wavelength:\[λ = \frac{h c}{U_{0}}\] where \(c\) is the speed of light \( = 3.00×10^8 m/s\).

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Key Concepts

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

Energy Levels
When studying the behavior of particles at the quantum level, we encounter distinct 'steps' or 'rungs' of energy, much like a ladder. These are known as the energy levels. In a quantum mechanical square well, a particle can only occupy these distinct energy states. The energy of each level is not continuous but quantized, meaning that the particle cannot possess energy in between these levels.

The formula for calculating the energy of a particle in a one-dimensional infinite potential well is given by \(E_n = \frac{n^2h^2}{8mL^2}\), where \(E_n\) represents the energy of the nth level, \(n\) is the principal quantum number, and only takes on positive integer values, \(h\) is Planck's constant, \(m\) is the mass of the particle (an electron in our case), and \(L\) is the width of the potential well. The square of \(n\) indicates that energy increases with the square of the principal quantum number — the higher the 'rung' on the 'ladder,' the greater the energy of the particle.
Ground State Energy
The ground state of a particle in a quantum mechanical square well is its lowest possible energy state, referred to as the baseline energy level. For the ground state, the principal quantum number \(n\) is equal to 1. Using the formula given for energy levels, we can deduce that \(E_1 = \frac{h^2}{8mL^2}\), since \(n = 1\). This represents the minimum energy the electron must have to be within the well.

Understanding ground state energy is crucial in quantum mechanics, as it forms the foundation for calculating excitation. Any energy added to the system will potentially 'lift' the electron to a higher energy level. The ability of an electron to exist at this basic level yet still be confined within the potential well underscores the fundamental principle of quantization in quantum mechanics.
Photon Absorption
Photon absorption is a process wherein an electron absorbs a photon, gaining its energy. This gain in energy can prompt the electron to jump to a higher energy level or, given enough energy, break free from the confines of the potential well entirely — a concept known as 'photoelectric effect' in a broader sense.

When an electron is bound within a potential well, it requires a precise amount of energy to escape, equal to the depth of the well, \(U_0\). The energy of a photon is determined by its wavelength and is given by the formula \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. To calculate the maximum wavelength of a photon that can still liberate an electron from the square well, we find the wavelength that corresponds with the energy equal to the well depth. Any longer wavelength would not have enough energy to free the electron, and thus, the electron will remain bound within the well.

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Most popular questions from this chapter

An electron with initial kinetic energy \(6.0 \mathrm{eV}\) encounters a barrier with height \(11.0 \mathrm{eV}\). What is the probability of tunneling if the width of the barrier is (a) \(0.80 \mathrm{nm}\) and (b) \(0.40 \mathrm{nm} ?\)

When low-energy electrons pass through an ionized gas, electrons of certain energies pass through the gas as if the gas atoms weren't there and thus have transmission coefficients (tunneling probabilities) \(T\) equal to unity. The gas ions can be modeled approximately as a rectangular barrier. The value of \(T=1\) occurs when an integral or half-integral number of de Broglie wavelengths of the electron as it passes over the barrier equal the width \(L\) of the barrier. You are planning an experiment to measure this effect. To assist you in designing the necessary apparatus, you estimate the electron energies \(E\) that will result in \(T=1\). You assume a barrier height of \(10 \mathrm{eV}\) and a width of \(1.8 \times 10^{-10} \mathrm{~m} .\) Calculate the three lowest values of \(E\) for which \(T=1\)

An electron is in a one-dimensional box. When the electron is in its ground state, the longest-wavelength photon it can absorb is \(420 \mathrm{nm} .\) What is the next longest-wavelength photon it can absorb, again starting in the ground state?

A particle of mass \(m\) in a one-dimensional box has the following wave function in the region \(x=0\) to \(x=L:\) $$\Psi(x, t)=\frac{1}{\sqrt{2}} \psi_{1}(x) e^{-i E_{1} t / \hbar}+\frac{1}{\sqrt{2}} \psi_{3}(x) e^{-i E_{3} t / \hbar}$$ Here \(\psi_{1}(x)\) and \(\psi_{3}(x)\) are the normalized stationary-state wave functions for the \(n=1\) and \(n=3\) levels, and \(E_{1}\) and \(E_{3}\) are the energies of these levels. The wave function is zero for \(x<0\) and for \(x>L\) (a) Find the value of the probability distribution function at \(x=L / 2\) as a function of time. (b) Find the angular frequency at which the probability distribution function oscillates.

The penetration distance \(\eta\) in a finite potential well is the distance at which the wave function has decreased to \(1 / e\) of the (b) wave function at the classical turning point: $$ \psi(x=L+\eta)=\frac{1}{e} \psi(L) $$ The penetration distance can be shown to be $$\eta=\frac{\hbar}{\sqrt{2 m\left(U_{0}-E\right)}}$$ The probability of finding the particle beyond the penetration distance is nearly zero. (a) Find \(\eta\) for an electron having a kinetic energy of \(13 \mathrm{eV}\) in a potential well with \(U_{0}=20 \mathrm{eV} .\) (b) Find \(\eta\) for a \(20.0 \mathrm{MeV}\) proton trapped in a 30.0 -MeV-deep potential well.

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