Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 12

Energy is required to compress the box when a particle is inside: this suggests that the particle exerts a force on the walls. (a) On the basis that when the length of the box changes by dL the energy changes by \(\mathrm{d} E=-F \mathrm{d} L,\) find an expression for the force. (b) At what length does \(F=1 \mathrm{N}\) when an electron is in the state \(n=1 ?\)

Short Answer

Expert verified
The force 'F' exerted by the particle on the box is given by \(F = n^2h^2/(4mL^3)\), and the length of the box 'L' when the force is 1 Newton is achieved when \(L = [(n^2h^2)/(4m)]^{1/3}\). For an electron in state \(n=1\), substitute the values for \(n, h\), and \(m\) into the expression to get 'L'.

Step by step solution

01

Rearrange the given equation to find 'F'

The given equation is already solved for 'F', so no rearrangement is needed. Therefore, the force 'F' is given by \(F = -\frac{dE}{dL}\). Note the negative sign, which denotes that the force applied by the particle against the walls of the box tends to expand it, or resist its compression.
02

Express total energy 'E' as a function of the box length 'L'

For the electron, according to the Schrödinger model, the total energy in state 'n' is known to be \(E = n^2 h^2/(8mL^2)\). Here, 'n' is the state of the electron, 'h' is Planck's constant, 'm' is the electron’s mass, and 'L' is the length of the box.
03

Differentiate the total energy 'E' with respect to 'L'

Apply the power rule in differentiation to find \(dE/dL\). The power rule states that if \(y = ax^n\), then \(dy/dx = nax^{n-1}\). Hence, \(dE/dL = -2n^2h^2/(8mL^3) = -n^2h^2/(4mL^3)\). Substitute the calculated value into the force formula from Step 1.
04

Find the length 'L' when force 'F' equals 1 N

To find the length 'L' when the force equals 1 Newton, substitute \(F = 1 N\) in the force formula from Step 3 and solve for 'L'. This equation becomes \(1 = n^2h^2/(4mL^3)\). Solving for 'L' gives \(L = [(n^2h^2)/(4m)]^{1/3}\).

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.

Start your free trial

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

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

Calculate the values of (a) \(\langle x\rangle\) (b) \(\left\langle x^{2}\right\rangle\) \(,(\mathrm{c})\left\langle p_{x}\right\rangle,(\mathrm{d})\left\langle p_{x}^{2}\right\rangle\) for a harmonic oscillator in its ground state by evaluation of the appropriate integrals (as in Problems \(2.13-2.15\) ). Examine the value of \(\Delta x \Delta p_{x}\) in the light of the uncertainty principle. Hint. Use the integrals \\[ \begin{array}{l} \int_{-\infty}^{\infty} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^{1 / 2} \\ \int_{0}^{\infty} x \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2 \alpha} \\\ \int_{-\infty}^{\infty} x^{2} \mathrm{e}^{-\alpha x^{2}} \mathrm{d} x=\frac{1}{2}\left(\frac{\pi}{\alpha^{3}}\right)^{1 / 2} \end{array} \\]

Demonstrate that accidental degeneracies can exist in a rectangular infinite square-well potential provided that the lengths of the sides are in a rational proportion, that is when \(L_{1}=\lambda L_{2},\) with \(\lambda\) an integer. What is the degeneracy?

(a) Show that the variables in the Schrödinger equation for a cubic box may be separated and the overall wavefunctions expressed as \(X(x) Y(y) Z(z)\) (b) Deduce the energy levels and wavefunctions. (c) Show that the function .are orthonormal. (d) What is the degeneracy of the level with \(E=14\left(h^{2} / 8 m L^{2}\right) ?\)

A particle was prepared travelling to the right with all momenta between \(\left(k-\frac{1}{2} \Delta k\right) \hbar\) and \(\left(k+\frac{1}{2} \Delta k\right) \hbar\) contributing equally to the wavepacket. Find the explicit form of the wavepacket at \(t=0,\) normalize it, and estimate the range of positions, \(\Delta x,\) within which the particle is likely to be found. Compare the last conclusion with a prediction based on the uncertainty principle. Hint. Use eqn 2.13 with \(g=B\) a constant, inside the range \(k-\frac{1}{2} \Delta k\) to \(k+\frac{1}{2} \Delta k\) and zero elsewhere, and eqn 2.12 with \(t=0\) for \(\Psi_{k} .\) To evaluate \(\int\left|\Psi_{k}\right|^{2} \mathrm{d} \tau\) (for the normalization step) use the integral \(\int_{-\infty}^{\infty}(\sin x / x)^{2} \mathrm{d} x=\pi .\) Take \(\Delta x\) to be determined (numerically) by the locations where \(|\Psi|^{2}\) falls to half its value at \(x=0\) For the last part use \(\Delta p_{x} \approx \hbar \Delta k\)

A particle of mass \(m\) is confined to a one-dimensional box of length \(L\). Calculate the probability of finding it in the following regions: (a) \(0 \leq x \leq \frac{1}{2} L,\) (b) \(0 \leq x \leq \frac{1}{4} L\) (c) \(\frac{1}{2} L-\delta x \leq x \leq \frac{1}{2} L+\delta x .\) Derive expressions for a general value of \(n\). Then evaluate the probabilities (i) for \(n=1\) (ii) in the limit \(n \rightarrow \infty\). Compare the latter to the classical expectations.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation
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