Concept explainers
(a)
To show that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a freely moving particle with the wave function
(a)
Answer to Problem 54P
It is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a freely moving particle with the wave function
Explanation of Solution
Write the Schrodinger’s equation.
Here,
Write the statement to be proved.
Here,
Write the expression of the given wavefunction.
Here,
Put equation (III) in equation (II).
Take the derivative equation (III) with respect to
Take the derivative of the above equation with respect to
Put equations (V) in the left-hand side of equation (II) and rearrange it.
Write the equation for the reduced Planck’s constant.
Here,
Write the equation for the wave vector.
Here,
Put equation (VII) and (VIII) in (VI).
Write the equation for the de Broglie wavelength.
Here,
Rewrite the above equation for
Put the above equation in equation (IX).
Write the equation for kinetic energy.
Put the above equation in equation (XI).
Conclusion:
Equation (XIII) is exactly the same as equation (IV) which has to be proved.
Thus, it is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a freely moving particle with the wave function
(b)
To show that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a particle in a box with the wave function
(b)
Answer to Problem 54P
It is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a particle in a box with the wave function
Explanation of Solution
Write the expression of the given wavefunction.
Put equation (XIV) in equation (II).
Take the derivative equation (XIV) with respect to
Take the derivative of the above equation with respect to
Put the above equation in the left-hand side of equation (XV) and rearrange it.
Put equation (VII) and (VIII) in the above equation.
Put equation (X) in the above equation.
Put equation (XII) in the above equation.
Conclusion:
Equation (XVI) is exactly the same as equation (XV) which has to be proved.
Thus, it is showed that the first term in the Schrodinger equation reduces to the kinetic energy of the quantum particle multiplies by the wavefunction for a particle in a box with the wave function
Want to see more full solutions like this?
Chapter 28 Solutions
Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term
- Verify that given by Equation 7.57 is a solution of Schrödinger's equation for the quantum harmonic oscillator. n(x)=Nne2x2/2Hn(x),n=0,1,2,3,...arrow_forwardA proton and a deuteron (which has the same charge as the proton but 2.0 times the mass) are incident on a barrier of thickness 11.8 fm and “height” 10.9 MeV. Each particle has a kinetic energy of 2.50 MeV. What is the ratio of the tunneling probability of the proton to the tunneling probability of the deuteron?arrow_forwardIf Ψ is the wave function, the probability density function is given by _____________ a) |Ψ| b) |Ψ|2 c) |Ψ|3 d) |Ψ|4arrow_forward
- A quantum particle is described by the wave function ψ(x) = A cos (2πx/L) for −L/4 ≤ x ≤ L/4 and ψ(x) everywhere else. Determine:arrow_forwardAn electron having total energy E = 4.50 eV approaches a rectangular energy barrier with U = 5.00 eV and L = 950 pm as shown in Figure P40.21. Classically, the electron cannot pass through the barrier because E < U. Quantum-mechanically, however, the probability of tunneling is not zero.(b) To what value would the width L of the potential barrier have to be increased for the chance of an incident 4.50-eV electron tunneling through the barrierto be one in one million?arrow_forwardShow that normalizing the particle-in-a-box wave function ψ_n (x)=A sin(nπx/L) gives A=√(2/L).arrow_forward
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax