Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the ratio of energy difference between the ground state and the first excited state for an infinite square well of length \(L\) to that of length \(2 L\). That is, find \(\left(E_{2}-E_{1}\right)_{L} /\left(E_{2}-E_{1}\right)_{2 L}\).

Short Answer

Expert verified
Answer: The ratio of the energy difference between the ground state and the first excited state for an infinite square well of length L to that of length 2L is 1/4.

Step by step solution

01

Find Energy of the Ground State and First Excited State for length L

To find the energy of the ground state (\(E_1\)) for length \(L\), we use the formula above with \(n = 1\). Similarly, to find the energy of the first excited state (\(E_2\)), we use \(n = 2\): $$E_{1}=\dfrac{1^2\hbar^2 \pi^2}{2mL^2}$$ $$E_{2}=\dfrac{2^2\hbar^2 \pi^2}{2mL^2}$$
02

Calculate the Energy Difference for Length L

Now that we have the energies of both states, the energy difference is given by: $$\Delta E_{L}=E_{2}-E_{1}$$ $$\Delta E_{L}=\dfrac{4\hbar^2 \pi^2}{2mL^2}-\dfrac{1\hbar^2 \pi^2}{2mL^2}$$ $$\Delta E_{L}=\dfrac{3\hbar^2 \pi^2}{2mL^2}$$
03

Find Energy of the Ground State and First Excited State for length 2L

Similar to Step 1, we find the energies for length \(2L\) as: $$E_{1}=\dfrac{1^2\hbar^2 \pi^2}{2m(2L)^2}$$ $$E_{2}=\dfrac{2^2\hbar^2 \pi^2}{2m(2L)^2}$$
04

Calculate the Energy Difference for Length 2L

Using the energies from Step 3, we calculate the energy difference: $$\Delta E_{2L}=E_{2}-E_{1}$$ $$\Delta E_{2L}=\dfrac{4\hbar^2 \pi^2}{2m(2L)^2}-\dfrac{1\hbar^2 \pi^2}{2m(2L)^2}$$ $$\Delta E_{2L}=\dfrac{3\hbar^2 \pi^2}{8mL^2}$$
05

Calculate the Desired Ratio

Finally, we calculate the ratio of the energy differences: $$\dfrac{\Delta E_{L}}{\Delta E_{2L}}=\dfrac{\frac{3\hbar^2 \pi^2}{2mL^2}}{\frac{3\hbar^2 \pi^2}{8mL^2}}$$ Using the fact that $ \dfrac{a}{c} / \dfrac{b}{c}=\dfrac{a}{b}$, $$\dfrac{\Delta E_{L}}{\Delta E_{2L}}=\dfrac{2mL^2}{8mL^2}=\dfrac{1}{4}$$ Therefore, the ratio of the energy difference between the ground state and the first excited state for an infinite square well of length \(L\) to that of length \(2L\) is \(\dfrac{1}{4}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Infinite Square Well
In quantum mechanics, the infinite square well is a model used to describe how particles, like electrons, behave in a confined space. Imagine a particle that is completely trapped in a box with walls so high that it cannot escape. This box is the infinite square well. The particle can move freely inside the box but cannot pass through the walls. This simple yet powerful concept helps us understand the quantum behavior of particles.

The infinite square well is often represented as a one-dimensional box with length \( L \). Because of the well's infinite depth, it imposes boundary conditions: the particle's wave function must go to zero at the walls.
  • This boundary condition results in only specific, quantized energies or states being allowed for the particle.
  • These energies depend on the width of the well \( L \) and the mass of the particle \( m \).
The energy levels in the well are described by a simple formula, where energy increases with the quantum number \( n \). By understanding these principles, students can explore more complex quantum systems.
Energy States
Energy states in quantum mechanics refer to the specific levels of energy that a quantum system can have. In the case of the infinite square well, these states are discrete, meaning they can only take on certain values. The ground state (\( E_1 \)) is the lowest energy state, and it occurs when the particle is in the least energetic arrangement possible.

As the particle moves to higher energy states, like the first excited state (\( E_2 \)), it occupies levels with increasing energy. The formula for the energy of these states is given by:
  • \( E_n = \frac{n^2 \hbar^2 \pi^2}{2mL^2} \)
Here, \( n \) is the quantum number, \( \hbar \) is the reduced Planck's constant, and \( m \) is the mass of the particle. What's interesting is:
  • As \( n \) increases, the particle's energy increases dramatically.
  • The energy levels are equidistant in the infinite square well.
This quantization is central to understanding quantum systems, as it contrasts with classical systems where energy is continuous.
Quantum Energy Levels
Quantum energy levels are the distinct energy values that a quantum system, like an electron in an infinite square well, can have. These levels are determined by the wave-like properties of particles described by quantum mechanics. In an infinite square well, energy levels are not influenced by external conditions but are entirely set by the well's size and the particle's mass.

Let's take a specific case to learn more. For a box of length \( L \), the energy difference between two consecutive energy levels, such as between the ground state and the first excited state, can be calculated using their individual energy formulas. The difference for length \( L \) is given by:
  • \( \Delta E_L = \frac{3\hbar^2 \pi^2}{2mL^2} \)
While for a box of double the length \( 2L \), it is:
  • \( \Delta E_{2L} = \frac{3\hbar^2 \pi^2}{8mL^2} \)
The ratio of these energy differences gives insight into how energy spacing changes with the size of the well. The ratio \( \frac{1}{4} \) suggests that doubling the well's length decreases the energy difference between states quite significantly. Understanding these concepts is crucial for grasping the vast and fascinating world of quantum mechanics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The ground state wave function for a harmonic oscillator is given by \(\Psi_{0}(x)=A_{2} e^{-x^{2} / 2 b^{2}}\). a) Determine the normalization constant \(A\). b) Determine the probability that a quantum harmonic oscillator in the \(n=0\) state will be found in the classically forbidden region.

The probability of finding an electron in a hydrogen atom is directly proportional to a) its energy. b) its momentum. c) its wave function. d) the square of its wave function. e) the product of the position coordinate and the square of the wave function. f) none of the above.

An approximate one-dimensional quantum well can be formed by surrounding a layer of GaAs with layers of \(\mathrm{Al}_{x} \mathrm{Ga}_{1-x}\) As. The GaAs layers can be fabricated in thicknesses that are integral multiples of the single-layer thickness, \(0.28 \mathrm{nm}\). Some electrons in the GaAs layer behave as if they were trapped in a box. For simplicity, treat the box as an infinite one-dimensional well and ignore the interactions between the electrons and the Ga and As atoms (such interactions are often accounted for by replacing the actual electron mass with an effective electron mass). Calculate the energy of the ground state in this well for these cases: a) 2 GaAs layers b) 5 GaAs layers

An electron is trapped in a one-dimensional infinite potential well that is \(L=300\). pm wide. What is the probability that one can detect the electron in the first excited state in an interval between \(x=0.500 L\) and \(x=0.750 L ?\)

Particle-antiparticle pairs are occasionally created out of empty space. Looking at energy-time uncertainty, how long would such particles be expected to exist if they are: a) an electron/positron pair? b) a proton/antiproton pair?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free