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Question #42861

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The Question

An electron of mass $m$ is confined in a one-dimensional infinite potential well of width $L$ between $x = 0$ and $x = L$ . The electron is in its first excited state ( $n=2$ ). Let $\Delta x$ , $\Delta p$ , and $\Delta E$ denote the standard deviations (uncertainties) in the position, momentum, and total energy of the electron in this state, respectively, and let $h$ be Planck's constant. A flawed argument suggests that because the probability density is exactly zero at the node $x = L/2$ , the electron is perfectly excluded from this point, implying a local spatial uncertainty $\Delta x \to 0$ and thus $\Delta E \to \infty$ . By correctly applying the principles of quantum mechanics to the entire stationary state, determine the true values of $\Delta x$ , $\Delta p$ , and $\Delta E$ .

Options

A.
$\Delta x = L\sqrt{\frac{1}{12} - \frac{1}{8\pi^2}}, \; \Delta p = \frac{h}{L}, \; \Delta E = 0$
B.
$\Delta x = 0, \; \Delta p \to \infty, \; \Delta E \to \infty$
C.
$\Delta x = L\sqrt{\frac{1}{12} - \frac{1}{2\pi^2}}, \; \Delta p = \frac{h}{2L}, \; \Delta E = \frac{3h^2}{8mL^2}$
D.
$\Delta x = L\sqrt{\frac{1}{12} - \frac{1}{8\pi^2}}, \; \Delta p = \frac{h}{L}, \; \Delta E = \frac{h^2}{2mL^2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Standing Wave Nodes/Antinodes|Heisenberg's Limitation|Energy Level Scaling

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental distinction between a mathematical node in a wavefunction and the physical localization of a particle.

Students often fall into the trap of assuming that because the probability density is zero at a node, the particle is 'perfectly excluded' from that exact point, leading them to incorrectly assign a local spatial uncertainty $\Delta x \to 0$ .

By misapplying Heisenberg's uncertainty principle to this single point, they deduce infinite momentum and energy uncertainties.

In reality, Heisenberg's principle applies to the standard deviations of the entire wave packet.

For a stationary state, the particle is delocalized across the entire potential well, yielding a finite $\Delta x$ and $\Delta p$ .

Furthermore, because it is an energy eigenstate, the energy is exactly defined, meaning $\Delta E = 0$ , regardless of the spatial nodal structure.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Waves - Class 11 Part 2
Concept	Standing Wave Nodes/Antinodes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free