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the question as mentioned in title. i want to know how this equation is dimenssionless as my efforts didn't give me the answer the equation is:

$$z=(L/ \hbar) \cdot \sqrt{2m \cdot (E+V_o)}$$

as i tried in draft i found :

[z]=m.Kg.s^-1

but in the solution sheet it's a dimensionless quantity. where z is the dimensionless quantity \hbar is 2* \pi * h (h: planck's constant) m is the mass of particle E and Vo are the energies of the system

mvw
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EnthusiastiC
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1 Answers1

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Note: It seems $L$ is not angular momentum, but some length.

For the SI units we have: $$ [L] = \text{m} \\ [\hbar] = \text{J}\text{s} = (\text{kg}\cdot (\text{m}/\text{s})^2) s = \text{m}^2 \cdot \text{kg} / \text{s} $$ And $$ [m] = \text{kg} \\ [E] = J = \text{kg}\cdot (\text{m}/\text{s})^2 \\ \left[\sqrt{m(E+V_0)}\right]= \text{kg}\cdot (\text{m}/\text{s}) = [p] $$ Combined we get: $$ \frac{\text{m}}{\text Js} \frac{\text{kg}\cdot\text{m}}{\text{s}} =\frac{\text{J}}{\text{J}} = 1 $$

mvw
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  • i think you made a mistake in the first line [L] has units of length [L]=m only so [L]/[\hbar] gives s/(m \cdot kg) when you multiply by the remainming term above \sqrt{2m \cdot (E+V_0)} the result appears as 1 – EnthusiastiC Apr 15 '16 at 22:14
  • Here we could be onto something. I considered $L$ to be angular momentum, but you say $L$ is meant as distance. – mvw Apr 15 '16 at 22:16
  • sorry i would be able to mention for this L: the width of finite square well anyway i think it's clear now since we have units of energy and length and momentum we can conclude it easily – EnthusiastiC Apr 15 '16 at 22:17