Quantum mechanics unit problem

[egslim]

I'm studying for a quantum mechanics test, but I've run into a problem with a practice sum I'm trying to solve.

The assignment is to calculate the energy in eV of the Lyman-alpha line. So I have to calculate E_n for n = 1.

That's easy with the formula for E_n which includes 4*pi*epsilon_0.

But I think I'm supposed to use this formula: E_n = (m*e^4)/(2*h_bar). Does anyone know what units I have to express m, e and h_bar in to find E_n = 13.6eV?

 

[cranreuch]

I'm studying for a quantum mechanics test, but I've run into a problem with a practice sum I'm trying to solve.

The assignment is to calculate the energy in eV of the Lyman-alpha line. So I have to calculate E_n for n = 1.

That's easy with the formula for E_n which includes 4*pi*epsilon_0.

But I think I'm supposed to use this formula: E_n = (m*e^4)/(2*h_bar). Does anyone know what units I have to express m, e and h_bar in to find E_n = 13.6eV?

Just figure it out in Joules using SI units and then convert to eV.
See Lyman_series on Wikipedia.

cranreuch

 

[Pragmatist]

I'm studying for a quantum mechanics test, but I've run into a problem with a practice sum I'm trying to solve.

The assignment is to calculate the energy in eV of the Lyman-alpha line. So I have to calculate E_n for n = 1.

That's easy with the formula for E_n which includes 4*pi*epsilon_0.

But I think I'm supposed to use this formula: E_n = (m*e^4)/(2*h_bar). Does anyone know what units I have to express m, e and h_bar in to find E_n = 13.6eV?

For that formula you should be using cgs units, but you have the formula slightly wrong it should be:

[latex]$$E_n = \frac{me^4}{2\hbar^2}$$[/latex]

Now, use e = 4.8032068(14) * 10^-10 esu, hbar = 1.05457266(63) * 10 ^-27 erg s and m in grams i.e.: 9.1093897(54) * 10^-28 g. This will give you the answer in ergs of roughly 2.1795 * 10^-11 ergs. Convert that to eV by multiplying by 6.2420 * 10^11 which is the number of eV in an erg, this will give you the 13.6 eV answer.

Alternatively you could use the SI formula and simply divide the answer in joules by the SI electronic charge (1.602 * 10^-19) and that will give you the same answer in eV as well.

 

[Schneibster]

Mmmfff. This is called Dimensional analysis^WP. It's a great way to check your answers.

First of all, the value of Planck's constant is
[latex]h =\,\,\, 6.626\ 068\ 96(33) \times10^{-34}\ \mbox{J}\cdot\mbox{s} \,\,\, = \,\,\, 4.135\ 667\ 33(10) \times10^{-15}\ \mbox{eV}\cdot\mbox{s}[/latex]
and the energy formula is
[latex]{E} = {h}\cdot{\nu}[/latex]
So we see that E, in joules, or eV, is equal to h, in joule-seconds, or electron-volt-seconds, times ν, in per-seconds. Notice that the seconds and the per-seconds cancel out, leaving the joules or the electron-volts.

Now Dirac's constant is also known as the "reduced" Planck constant, and it is
[latex]\hbar\ \equiv \frac{h}{2\pi} = \,\,\, 1.054\ 571\ 628(53)\times10^{-34}\ \mbox{J}\cdot\mbox{s} \,\,\, = \,\,\, 6.582\ 118\ 99(16) \times10^{-16}\ \mbox{eV}\cdot\mbox{s}[/latex]
Notice here that both 2 and pi are dimensionless; they contribute nothing to the dimensional analysis. So the units of h_bar are the same as the units of h: joule-seconds or electron-volt-seconds.

So if your formula is
[latex]{E_n} = \frac{{m}{e^4}}{{2}{\hbar}}[/latex]
then you have a problem, because the dimensional analysis doesn't work out. The REAL formula is
[latex]E_n = - {{m e^4} \over {2 \left( 4 \pi \varepsilon_0 \hbar \right)^2}} {1 \over n^2} = - {13.6 \over n^2} [\mbox{eV}][/latex]
which for n = 1 will be 13.6eV. Note the permittivity of the vacuum in the denominator, and the fact that it and the Dirac constant are squared. Dimensional analysis will tell you the correct formula, if you listen to it.

 

[Paul C. Anagnostopoulos]

I just finished producing a book on quantum chemistry, so I went to check Schneibster's final equation. The author expressed it as

[latex]$E_n = -\frac{m_{\mathrm{e}}e^4}{32\pi^2\epsilon_0^2\hbar^2n^2}$[/latex]

so Schneibster is correct.

~~ Paul

 

[Pragmatist]

Mmmfff. This is called Dimensional analysis^WP. It's a great way to check your answers.

First of all, the value of Planck's constant is
[latex]h =\,\,\, 6.626\ 068\ 96(33) \times10^{-34}\ \mbox{J}\cdot\mbox{s} \,\,\, = \,\,\, 4.135\ 667\ 33(10) \times10^{-15}\ \mbox{eV}\cdot\mbox{s}[/latex]
and the energy formula is
[latex]{E} = {h}\cdot{\nu}[/latex]
So we see that E, in joules, or eV, is equal to h, in joule-seconds, or electron-volt-seconds, times ν, in per-seconds. Notice that the seconds and the per-seconds cancel out, leaving the joules or the electron-volts.

Now Dirac's constant is also known as the "reduced" Planck constant, and it is
[latex]\hbar\ \equiv \frac{h}{2\pi} = \,\,\, 1.054\ 571\ 628(53)\times10^{-34}\ \mbox{J}\cdot\mbox{s} \,\,\, = \,\,\, 6.582\ 118\ 99(16) \times10^{-16}\ \mbox{eV}\cdot\mbox{s}[/latex]
Notice here that both 2 and pi are dimensionless; they contribute nothing to the dimensional analysis. So the units of h_bar are the same as the units of h: joule-seconds or electron-volt-seconds.

So if your formula is
[latex]{E_n} = \frac{{m}{e^4}}{{2}{\hbar}}[/latex]
then you have a problem, because the dimensional analysis doesn't work out. The REAL formula is
[latex]E_n = - {{m e^4} \over {2 \left( 4 \pi \varepsilon_0 \hbar \right)^2}} {1 \over n^2} = - {13.6 \over n^2} [\mbox{eV}][/latex]
which for n = 1 will be 13.6eV. Note the permittivity of the vacuum in the denominator, and the fact that it and the Dirac constant are squared. Dimensional analysis will tell you the correct formula, if you listen to it.

And in which unit system exactly does one use a value of h or hbar in eV s ? Certainly not the one you claim is the "REAL" equation...

There is no "REAL" equation per se. The equation you have given is correct for an input of quantities in SI units and an output in SI units (i.e. Joules). If you want the "REAL" equation for an SI input and an eV output then it should be:

[latex]$$\frac{me^3}{32 \pi^2 \varepsilon_0^2 \hbar^2 n^2}$$[/latex]

The one you have stated is dimensionally incorrect for eV which should have shown up if you'd actually done the dimensional analysis...

And in this case the n is redundant because egslim made it clear he is calculating the Lyman alpha.

However, problems like this are often stated in cgs units in academic tests because those units are more convenient for research and theoretical work (avoiding unnecessary force constants and "permittivities" etc.) So the correct form of the equation in cgs is as I stated it:

[latex]$$\frac{me^4}{2\hbar^2}$$[/latex]

for an output in ergs. Which is much simpler to write. It then only needs an eV conversion. So egslim was correct except that he missed the square from the hbar. It is not a question that there is only one "REAL" equation. And a dimensional analysis should work out properly regardless of which unit system is used, provided the units themselves are consistent.

 

[Pragmatist]

I just finished producing a book on quantum chemistry, so I went to check Schneibster's final equation. The author expressed it as

[latex]$E_n = -\frac{m_{\mathrm{e}}e^4}{32\pi^2\epsilon_0^2\hbar^2n^2}$[/latex]

so Schneibster is correct.

~~ Paul

That formula is correct for SI units so it gives an answer in joules not eV. If you want the answer in eV it needs to be e^3 not e^4 in the numerator. The problem is that many courses are still taught in cgs units because the formulae are simpler and easier to remember (although the unit conversions and dimensions are not so straightforward).

 

[Schneibster]

Heh, Pragmatist screws up another newb. Nice one, Pragmatist. Considering the textbooks and references on-line say you're wrong, and considering there's a direct conversion from eV to joules to ergs, what precisely are you talking about? You just said that the dimensional analysis is different if you switch systems of units that convert directly into one another. Do you even comprehend why this is wrong?

Lemme ask you something: do you actually know anything about physics, or do you just post stuff at random because I showed you didn't know what you were talking about, so you can prove I'm "wrong?" That's all I've ever seen you do here.

 

[egslim]

For that formula you should be using cgs units, but you have the formula slightly wrong it should be:

[latex]$$E_n = \frac{me^4}{2\hbar^2}$$[/latex]

Now, use e = 4.8032068(14) * 10^-10 esu, hbar = 1.05457266(63) * 10 ^-27 erg s and m in grams i.e.: 9.1093897(54) * 10^-28 g. This will give you the answer in ergs of roughly 2.1795 * 10^-11 ergs. Convert that to eV by multiplying by 6.2420 * 10^11 which is the number of eV in an erg, this will give you the 13.6 eV answer.

Great, thanks!

I tried something similar, but forgot to convert the ergs to eV, silly me.

Just figure it out in Joules using SI units and then convert to eV.

That means I'd have to memorize several conversion factors, since they probably won't be supplied for the test. I'm very happy that's no longer necessary.

The REAL formula is

As I mentioned in the first post, though not very clearly, when using the "real formula" I had no difficulty obtaining the right answer. However, the textbook for this course uses Pragmatists formula. Converting the variables supplied in the text to SI (without a conversion table, since we're not allowed to bring books) would be hopelessly impractable.

Thanks, everyone!

 

[Schneibster]

The REAL formula is
[latex]E_n = - {{m e^4} \over {2 \left( 4 \pi \varepsilon_0 \hbar \right)^2}} {1 \over n^2} = - {13.6 \over n^2} [\mbox{eV}][/latex]
which for n = 1 will be 13.6eV. Note the permittivity of the vacuum in the denominator, and the fact that it and the Dirac constant are squared. Dimensional analysis will tell you the correct formula, if you listen to it.

So since Pragmatist has muddied the waters, let's do dimensional analysis on this equation.

On the left side, we have energy. So the right side must put out energy; I will stick with joules, but remember that you can convert to eV or ergs, and the dimensional analysis will still give the correct answer.

In the numerator, we have the mass of the electron. We also have the charge of the electron, to the fourth power. Thus we have
kg x C^4

In the denominator, we have Dirac's constant, in j x s, and this is squared so we have j^2 x s^2. We also have the permittivity of the vacuum, which is in C^2 / (N x m^2). This is also squared, so we have C^4 / (N^2 x m^4). The whole units of the denominator are therefore
(j^2 x s^2 x C^4)/(N^2 x m^4).

Cancelling charge, since it is ^4 in both the numerator and denominator, we are left with
kg / (j^2 x s^2)/(N^2 x m^4)

Now, joules are newton-meters. So we can say that we have
kg / (N^2 x m^2 x s^2)/(N^2 x m^4)
Reducing,
kg / (s^2)/(m^2)
or
kg x m^2/s^2
Now, joules are also kg x m^2/s^2, so we see that we have joules on the right side.

Thus, we have j = j.

 

[Schneibster]

BTW, if that's what's in your textbook, you better use it, but remember this: it's not right. You'll apparently find out about that later. Dimensional analysis always tells the truth.

 

[Pragmatist]

Heh, Pragmatist screws up another newb. Nice one, Pragmatist.

Go ahead, insult the questioner, why don't you?

Considering the textbooks and references on-line say you're wrong,

Oh they do, do they? Which "text books" are these exactly? This isn't one of your amazing self written or non-existent Wikipedia references again is it?

and considering there's a direct conversion from eV to joules to ergs, what precisely are you talking about? You just said that the dimensional analysis is different if you switch systems of units that convert directly into one another. Do you even comprehend why this is wrong?

Since you seem determined to dig yourself into a deeper hole as usual, please explain how systems of units "directly convert into one another". Which systems of units? The cgs system has totally different dimensions to SI, there is no "direct" conversion - a conversion always involves a change of actual dimensions in some of the quantities. For example cgs doesn't use permittivities etc., so where do the units of the permittivity go in cgs if there is a "direct" conversion?

As for the rest, egslim asked how to calculate the value of the Lyman alpha. He did not ask how to calculate the Lyman beta, gamma etc. Furthermore, he did not ask for the ionization energy of Hydrogen (which is numerically equal but is technically a different thing entirely), so quoting him a negative energy as you did does not give the alpha photon, it gives the ionization energy. But since you're always right, please feel free to go ahead and explain how you ended up with a negative energy photon. This should be amusing!

Lemme ask you something: do you actually know anything about physics, or do you just post stuff at random because I showed you didn't know what you were talking about, so you can prove I'm "wrong?" That's all I've ever seen you do here.

I know more about it than you do as is obvious from the numerous threads where you give misinformation and end up throwing insults at everyone who tries to correct you - not to mention how you have behaved numerous times when proven wrong. And you haven't "shown" anything (except your usual ignorance and poor grace) your claims do not consititute proofs.

 

[Pragmatist]

So since Pragmatist has muddied the waters, let's do dimensional analysis on this equation.

On the left side, we have energy. So the right side must put out energy; I will stick with joules, but remember that you can convert to eV or ergs, and the dimensional analysis will still give the correct answer.

In the numerator, we have the mass of the electron. We also have the charge of the electron, to the fourth power. Thus we have
kg x C^4

In the denominator, we have Dirac's constant, in j x s, and this is squared so we have j^2 x s^2. We also have the permittivity of the vacuum, which is in C^2 / (N x m^2). This is also squared, so we have C^4 / (N^2 x m^4). The whole units of the denominator are therefore
(j^2 x s^2 x C^4)/(N^2 x m^4).

Cancelling charge, since it is ^4 in both the numerator and denominator, we are left with
kg / (j^2 x s^2)/(N^2 x m^4)

Now, joules are newton-meters. So we can say that we have
kg / (N^2 x m^2 x s^2)/(N^2 x m^4)
Reducing,
kg / (s^2)/(m^2)
or
kg x m^2/s^2
Now, joules are also kg x m^2/s^2, so we see that we have joules on the right side.

Thus, we have j = j.

Which is what I said all along. The answer was supposed to be in eV and you claimed before that the answer was in eV, now you show it is in joules...

Doesn't your foot hurt when you shoot holes in it?

 

[Pragmatist]

Great, thanks!

I tried something similar, but forgot to convert the ergs to eV, silly me.

You're welcome!

 

[Yllanes]

That means I'd have to memorize several conversion factors, since they probably won't be supplied for the test. I'm very happy that's no longer necessary.

The conversion factor for the change eV <-> J is, numerically, the electron charge in C (this is obvious from the very definition of eV). So it's not an apparently obscure conversion factor like, for example, the one between esu and C, but a fundamental constant you have to know anyway. Also, a very important number to keep in mind is

[latex]\footnotesize
\[
\alpha = \frac{e^2}{[4\pi\epsilon_0]\hbar c} = \frac{1}{137.036}
\]
[/latex]

(valid for SI units, remove the factors in brackets to use CGS units, remove hbar and/or c if you are using natural units). It is profitable to know this definition, you will find yourself recurring to it more often as you progress on to more advanced topics. In nuclear physics, for example, the only two constants one really has to remember are that one and hbar · c = 197.3 MeV·fm.

 

[Pragmatist]

To follow up on what Yllanes just said, alpha can be very useful. For example one of the simplest ways of expressing Hydrogen like energy levels is:

[latex]$$E_n = \frac{E_e\alpha^2}{2n^2}$$[/latex]

Where [latex]$E_e$[/latex] is the intrinsic rest energy of the electron i.e.:

[latex]$$E_e = m_e c^2$$[/latex]

 

[Yllanes]

To follow up on what Yllanes just said, alpha can be very useful. For example one of the simplest ways of expressing Hydrogen like energy levels is:

[latex]$$E_n = \frac{E_e\alpha^2}{2n^2}$$[/latex]

I prefer

[latex]\footnotesize
\[
E_n = -\frac12 m_e (\alpha Z c)^2\frac{1}{n^2}
\]
[/latex]

I think it is easier to remember, if you consider the electron to have an average speed of alpha*c (which is true in this model). This way the formula looks like a kinetic energy. Z is the charge of the nucleus (the formula works for ionised atoms).

 

[Paul C. Anagnostopoulos]

Note to self: Stay the hell out of mathematical physics threads, except as a casual observer.

~~ Paul

 

[andyandy]

I have a question.....

If we measure the heat transfer rate as [latex]${\dot Q}[/latex] then how many 40 watt light bulbs can an innocuous physics question power through the heat generated on it?




*don't correct my physics *

 

[Paul C. Anagnostopoulos]

That formula is correct for SI units so it gives an answer in joules not eV. If you want the answer in eV it needs to be e^3 not e^4 in the numerator. The problem is that many courses are still taught in cgs units because the formulae are simpler and easier to remember (although the unit conversions and dimensions are not so straightforward).

Is the eV considered part of the CGS system? I thought it was sort of free floating, not part of either system.

~~ Paul