How to deduce the units of hw?

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In summary: Yes, i am sure, the m0 is in the right units. Thanks for the help.In summary, Harrison's book states that Planck's constant, in those units, is actually 6.5821... \times 10^{-13} meV s. There is a 'seconds' in Planck's constant. Thus the angular frequency is 1.325 \times 10^{15} {{radians}\over{seconds}}.
  • #1
cashimingo
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Hi everyone, in the book of Paul Harrison (Quantum Wells, Wires and Dots, 2nd Ed.), it is said that [itex]\hbarω=871.879 meV[/itex] but it is known that [itex]\hbar=6.58214928\times10^{-13}meV[/itex] so i can inferred that [itex]\omega=9.9879927[/itex], but i don't know the units of [itex]\omega[/itex], or how can i deduce what units he used for [itex]\omega[/itex] and what units for [itex]\hbar[/itex], so that when i multiply both, it results 871.879 meV?.

Any suggestion would be really appreciated.
Thanks.
 
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  • #2
Planck's constant, in those units, is actually [itex]\hbar = 6.5821... \times 10^{-13} meV s[/itex]. There is a 'seconds' in Planck's constant. Thus the angular frequency is [itex]1.325 \times 10^{15} {{radians}\over{seconds}}[/itex].
 
  • #3
Ah ok, yeah, but it is also true that [itex]\hbar=87.2827...[/itex] but in what units?.
Because (87.2827..)(9.9878..)=871.879 meV, my problem is that i don't know what are the units of each number.

Thanks for the reply.
 
  • #4
cashimingo said:
Ah ok, yeah, but it is also true that [itext]\hbar=87.2827...[\itext] but in what units?.
Because (87.2827..)(9.9878..)=871.879 meV, my problem is that i don't know what are the units of each number.

Thanks for the reply.

You can arbitrarily make up units to make Planck's constant whatever you want, but I've never seen Planck's constant with the numerical value of 87.2827...

The common values for Planck's constant is given at http://en.wikipedia.org/wiki/Planck's_constant .
 
  • #5
Yes that's true, that is what I'm trying to figure out? in what units is the 87.2927...?,
 
  • #6
cashimingo said:
Yes that's true, that is what I'm trying to figure out? in what units is the 87.2927...?,

You did your calculation incorrectly at the very start. You should have determined that [itex]\omega = 1.32\times 10^{15} {{radians}\over{seconds}}[/itex]. No SI units would allow you to get the 87.2927 number for Planck's constant.
 
  • #7
Well, maybe i haven't explain the problem correctly. In Harrison's book, there is the following calculation

[itex]\omega=\sqrt{8\Delta V_{CB}(E_{g}^{AlAs}-E_{g}^{GaAs})(x_{max}-x_{min})} \frac{1}{a\sqrt{m}} [/itex]

where
[itex]V_{CB}=0.67,[/itex]
[itex]E_{g}^{AlAs}-E_{g}^{GaAs}=1247 meV,[/itex]
[itex]x_{max}=10,[/itex]
[itex] x_{min}=0,[/itex]
[itex]a=100angstroms [/itex]
[itex]m=0.067 m_{0}[/itex]

then (that's what he says) [itex]\hbar\omega=871.879 meV[/itex].
But when i perform the calculation, i get
[itex]\omega≈9.9879927[/itex], so i get [itex]\hbar≈87.2927[/itex].
That's why i asked in what units was the [itex]\hbar[/itex] and the [itex]\omega[/itex].

Any comment of the calculation or what is wrong, would be really appreciated.
Thanks.
By the way, how can i write the angstrom symbol?
 
  • #8
Cashmingo pengwuino has answered your question.h bar has units of meVs and not meV as you wrote incorrectly in post 1.Omega has units of seconds to the minus one and so,therefore,h bar omega has units of meV as you wrote correctly in post 1.
 
  • #9
cashimingo said:
But when i perform the calculation, i get
[itex]\omega≈9.9879927[/itex]

That is off by roughly 14 orders of magnitude. I crosschecked and got [itex]\omega≈1.325*10^{15}\frac{1}{s}[/itex] which matches the energy given in the book.

Are you sure you used the right units for m0? It is [itex]\frac{510998.928 eV}{c^2} [/itex].
 

FAQ: How to deduce the units of hw?

1. How do I determine the units of hw?

The units of hw can be determined by looking at the units of the given variables in the equation and using dimensional analysis to cancel out units until only the desired units remain.

2. Is there a specific formula for deducing the units of hw?

There is no specific formula for deducing the units of hw as it depends on the specific variables and equations being used. However, dimensional analysis is a commonly used method for determining units.

3. Can I use any units for hw?

No, the units of hw must be consistent with the units of the variables in the equation. For example, if the variables are measured in meters and seconds, the units of hw must also be in meters.

4. Are there any shortcuts for deducing the units of hw?

While there are no shortcuts for determining the units of hw, it can be helpful to break down the equation into smaller parts and determine the units for each part separately before combining them.

5. How important is it to correctly deduce the units of hw?

Deducing the units of hw is crucial in order to ensure that the final result is in the correct units and is mathematically accurate. It also helps to identify any errors or inconsistencies in the equation or calculations.

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