# Find n of a mass confined between 2 rigid walls?

## Homework Statement

A small object of mass 1mg is confined to move between two rigid walls separated by 1cm, if the speed of the object is 3cm/s find the corresponding value of n

## Homework Equations

$$P=\sqrt{2mE}$$
$$E = \frac{h^{2}}{8mL^{2}}$$

## The Attempt at a Solution

V = 3cm/s = .03m/s
m = 1mg = $$1x10^{-6}kg$$

P = mv
$$mv=\sqrt{2mE}$$
$$(mv)^{2} = 2mE$$
$$E=\frac{(mv)^{2}}{2m} = \frac{[(.03)(1x10^{-6})]^{2}}{2} = 4.5x10^{-16}J$$

$$E = \frac{h^{2}}{8mL^{2}}$$

$$n = \sqrt{\frac{8EmL^{2}}{h^{2}}}$$

$$= \sqrt{\frac{(8)(4.5x10^{-16})(.01)(1x10^{-6})^{2}}{(6.626x10^{-34})^{2}}}$$

I get...$$9.1x10^{18}$$

seems rather large...any ideas?

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Shooting Star
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I get...$$9.1x10^{18}$$

seems rather large...any ideas?
First of all, your calculation is probably not correct. (But you check it once more.)

Since n is dimensionless, it can be evaluated in any consistent system of units. There’s no need to introduce the momentum p etc. and make the thing unnecessarily lengthy and messy.

Using the CGS system, where h=6.63*10^-27 erg-sec, and writing E = ½ mv^2,

$$n = \sqrt{8EmL^2/h^2} = \sqrt{{8(1/2)mv^2*L^2}/{h^2}} = 2mvL/h$$ (becomes so simple!)

$$= 2*10^{-3}*3*1/(6.63*10^{-27}) = 9*10^{23}$$. (Please check once.)

But this n is also high, meaning that the particle behaves like a classical particle within the box, spending more or less equal times in all parts of the box, because there is a very large number of half waves between the two walls.

If n was 1, say, there would be only one half wave, which would mean that the particle would be spending more time near the middle of the box, since the amplitude of the standing wave and thus it’s square is higher near the middle, and so the probability of finding it near the middle is higher than at other points.

This is an example of the Correspondence Principle, which states that at high quantum numbers, quantum mechanical results tend to reduce to classical mechanical ones.