(a)
The potential energy of the system.
(a)
Answer to Problem 44CP
The potential energy of the system is
Explanation of Solution
Write the equation for potential energy.
Here,
Substitute
Here,
Write the equation for the potential energy of the system.
Here,
Conclusion:
Substitute equation (II) in equation (III) for each pair of energy.
Thus, the potential energy of the system is
(b)
The minimum kinetic energy of the two electrons.
(b)
Answer to Problem 44CP
The minimum kinetic energy of the two electrons is
Explanation of Solution
Let the minimum energy of the electron be
Here,
Substitute
Write the equation for the minimum kinetic energy of the two electrons.
Here,
Conclusion:
Substitute equation (V) in equation (VI) to find
Thus, the minimum kinetic energy of the two electrons is
(c)
The value of
(c)
Answer to Problem 44CP
The value of
Explanation of Solution
Write the equation for the total energy.
Here,
Substitute equation (IV) and (VII) in the above equation to find
Write the condition for the energy to be minimum.
Conclusion:
Substitute equation (VIII) in the above equation to find
Rearrange the above equation for
Substitute
Thus, the value of
(d)
Compare the value of
(d)
Answer to Problem 44CP
The lithium inter-atomic spacing is in the same order of magnitude as the interatomic spacing of
Explanation of Solution
Write the equation for volume.
Here,
Write the equation for density.
Here,
Substitute equation (X) in the above equation and rearrange to find
Conclusion:
Substitute
The lithium interatomic spacing is 5.59 times larger than
Thus, the lithium interatomic spacing is in the same order of magnitude as the interatomic spacing
Want to see more full solutions like this?
Chapter 40 Solutions
Physics for Scientists and Engineers with Modern Physics
- Consider hydrogen in the ground state, 100 . (a) Use the derivative to determine the radial position for which the probability density, P(r), is a maximum. (b) Use the integral concept to determine the average radial position. (This is called the expectation value of the electrons radial position.) Express your answers into terms of the Bohr radius, a0. Hint: The expectation value is the just average value, (c) Why are these values different?arrow_forward(a) “An electron is caught by the interatomic void site of a Cu crystal and exhibits 0.3 and 0.7 probabilities at a time in the middle point of the trap” Discuss the validity of this observation and explain why.(b) “The state function of a hydrogen atom is broken at a space point on its way towards it mean position when the atom is oscillating” Justify the possibility of this.arrow_forwardWhen an atom drops from an initial level to a lower energy level or is raised from an initial level to a higher energy level, what should be the value of the energy of the photon for this to situation to happen? a. twice the energy gap between the ground and excited states b. equal to the energy gap between the ground and excited states c. less than the energy gap between the ground and excited states d. greater than the energy gap between the ground and excited statesarrow_forward
- A laser medium is confined to a cavity that ensures that only certain photons of a particular frequency, direction of travel, and state of polarization are generated abundantly. The cavity is essentially a region between two mirrors, which reflect the light back and forth. This arrangement can be regarded as a version of the particle in a box, with the particle now being a photon. As in the treatment of a particle in a box (Topic 7D), the only wavelengths that can be sustained satisfy n × 1/2λ = L, where n is an integer and L is the length of the cavity. That is, only an integral number of half-wavelengths fit into the cavity; all other waves undergo destructive interference with themselves. These wavelengths characterize the resonant modes of the laser. For a laser cavity of length 1.00 m, calculate (a) the allowed frequencies and (b) the frequency difference between successive resonant modes.arrow_forwardA 5.0-eV electron impacts on a barrier of with 0.60 nm. Find the probability of the electron to tunnel through the barrier if the barrier height is (a) 7.0 eV; (b) 9.0 eV; and (c) 13.0 eV.arrow_forwardA 6.0-eV electron impacts on a barrier with height 11.0 eV. Find the probability of the electron to tunnel through the barrier if the barrier width is (a) 0.80 nm and (b) 0.40 nm.arrow_forward
- A quantum particle with initial kinetic energy 32.0 ev encounters a square barrier with height 41.0 ev and width 0.25 nm. Find probability that the particle tunnels through this barrier if the particle is (a) an electron and, (b) a proton.arrow_forwardAn electron is trapped in a one-dimensional infinite potential well. For what (a) higher quantum number and (b) lower quantum number is the corresponding energy difference equal to the energy difference E43 between the levels n = 4 and n = 3? (c) Show that no pair of adjacent levels has an energy difference equal to 2E43.arrow_forwardSome of the most powerful lasers are based on the energy levels of neodymium in solids, such as glass, as shown in the figure. Verify that the 1.17 eV transition produces 1.06 µm radiation: Fill in the blanks for the work done to solve ___a)__________ev m ? = ___________________________________ = 1.06 µm ___b)__________evarrow_forward
- wave function of hydrogen atom is given by sie = sie100+ sie210+3sie211, find the expectation value of L^2 in state of siearrow_forwardQ. (a) Calculate the zero point of an electron in a cubic box of edge-length 2.0 nmarrow_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_forward
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningModern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning