Potential energy, Wave function, Quantum physics

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The discussion centers around confusion regarding a task related to potential energy, wave functions, and quantum physics. Participants suggest that a foundational understanding of the relevant material, including solutions and equations, is necessary to tackle the task effectively. Clarification on specific concepts or equations may be needed to assist with the confusion. Engaging with study materials or lectures is recommended for better comprehension. Overall, a grasp of the underlying principles of quantum physics is essential for addressing the task at hand.
HektorHusky
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Homework Statement
A one-dimensional potential well has a barrier of height 1.5 eV (relative to the energy of the bottom of the well) on the right hand side, and a barrier higher than this on the left hand side. We happen to know that this potential well has an energy eigenstate for an electron at 1.3 eV (also relative to the energy at the bottom of the well)
Relevant Equations
State the general form of the wavefunction solution (i.e., within a normalizing constant that you need not attempt to determine) in each of the following two cases, giving actual values for any wavevector magnitude k and/or decay constant k in these wavefunctions. i. Within the well. ii. In the barrier on the right hand side.
I am totally confused about the task. Any help will be nice. Thank you so much
 
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I assume you have been studying or attending lectures on the relevant material? The question appears to assume you have a whole bunch of solutions and equations at your fingertips.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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