Particle in a box

A particle of mass m m is confined between two walls, separated by a distance L . L. The particle is in the lowest energy state (ground state), and when the separation of the walls is L 0 L_0 its energy is E 0 = h 2 8 m L 0 2 , E_0= \frac{h^2}{8mL_0^2}, where h h is the Planck's constant.

The particle exerts a force F F on the walls that is trying to push the walls apart. If we change the distance between the walls, the force is changing, and it can be expressed as

F = 2 E 0 L ( L 0 L ) A . F= \frac{2E_0}{L}\left(\frac{L_0}{L}\right)^A.

What is the value of A ? A?


The answer is 2.0.

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1 solution

Laszlo Mihaly
Oct 1, 2018

The force is F = d E d L F=-\frac{dE}{dL} where the energy is E = n h 2 8 m L 2 E=\frac{nh^2}{8mL^2} and in the ground state n = 1 n=1 . Accordingly

F = h 2 8 m 2 1 L 3 = 2 E 0 L ( L 0 L ) 2 F= \frac{h^2}{8m} 2\frac{1}{L^3}= \frac {2E_0}{L} \left(\frac{L_0}{L}\right)^2

and therefore A = 2 A=2 .

Interestingly, the answer is the same for a classical (no quantum mechanics) particle, see here , although the calculation is entirely different in the two problems.

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