Difference in Physics and Chemistry Text regarding Wavenumber

[maverick280857]

Hi

Those of you who have read Bohr's Theory in Chemistry may have encountered the relation,

<br /> \frac{1}{\lambda} = RhcZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})<br />

for the wavelength of radiation emitted when an electron goes from a higher energy level n_{2} to a lower energy level n_{1}, R is the Rydberg Constant, c is the speed of light and Z is the atomic number of the one-electron (hydrogen-like) species being considered.

Now some books refer to the fraction \frac{1}{\lambda} as the "wavenumber", whereas in physics, the fraction \frac{2\pi}{\lambda} is called the wavenumber. Why should this difference exist at all?

I was told by my teachers to make a distinction when answering questions on physics (use the second formula) and chemistry (use the first one) but that to me seems hardly convincing.

Cheers
Vivek

 

[selfAdjoint]

Hi

Those of you who have read Bohr's Theory in Chemistry may have encountered the relation,

<br /> \frac{1}{\lambda} = RhcZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})<br />

for the wavelength of radiation emitted when an electron goes from a higher energy level n_{2} to a lower energy level n_{1}, R is the Rydberg Constant, c is the speed of light and Z is the atomic number of the one-electron (hydrogen-like) species being considered.

Now some books refer to the fraction \frac{1}{\lambda} as the "wavenumber", whereas in physics, the fraction \frac{2\pi}{\lambda} is called the wavenumber. Why should this difference exist at all?

I was told by my teachers to make a distinction when answering questions on physics (use the second formula) and chemistry (use the first one) but that to me seems hardly convincing.

Cheers
Vivek

I believe it's because Planck's constant h comes into Rydberg's constant and in its original form. But physicists are accustomed to use \hbar = h/{2\pi}. So when physicists use Rydberg's constant, they have to divide it by 2\pi.

 

[Gokul43201]

Also, physicists use the angular frequency \omega = 2 \pi f more often than the regular frequency, f.

So ~\omega = 2\pi f = 2 \pi \frac {c} {\lambda} = c \frac {2\pi} {\lambda}

Hence the popularity of that form among physicists.

 

[maverick280857]

Thanks selfAdjoint and Gokul43201

I guess I do have to make this distinction while doing physics and chemistry :-D

Cheers
Vivek

 

[maverick280857]

Just to add to what I posted earlier, I learned the following from my teacher very recently:

The wavenumber as defined by the Ritz Formula, which is,

\frac{1}{\lambda} = RZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}) \qquad (n_{2} &gt; n_{1})<br />

(and not what I had mentioned in my first mail)

is just the reciprocal of the wavelength (which should mean in physical terms, the number of oscillations per unit length) but in Bohr's terms, when the wave is "fit" into the "orbit", the wavenumber concerned with the stationary wave is indeed the one that comes from physics, that is \frac{2\pi}{\lambda}.

So its just that these two quantities with similar names must be distinguished in the context of their usage. If they ask you for the wavenumber of the alpha line in the Balmer spectral series of an unielectron species then you have to use the Ritz relationship setting n_{1} = 2 and n_{2} = 3, but if the wavenumber is referred to in terms of the stationary wave that fits into the Bohr Orbit (the so called "de-Broglie wave") then you have to use \frac{2\pi}{\lambda}.

Cheers
Vivek

 

[Gokul43201]

You can write down either 1/ \lambda ~~or~~2\pi / \lambda as long as you provide the correct units to avoid confusion - m^-1 for the former and rad/m for the latter.