Difference in Physics and Chemistry Text regarding Wavenumber

In summary, the wavelength of radiation emitted when an electron goes from a higher energy level to a lower energy level is determined by the Rydberg Constant, the speed of light, and the atomic number of the one-electron (hydrogen-like) species being considered. The wavenumber as defined by the Ritz Formula is different than the wavenumber used in physics, which is determined by the stationary wave that fits into the Bohr Orbit.
  • #1
maverick280857
1,789
4
Hi

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

[tex]
\frac{1}{\lambda} = RhcZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})
[/tex]

for the wavelength of radiation emitted when an electron goes from a higher energy level [tex]n_{2}[/tex] to a lower energy level [tex]n_{1}[/tex], 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 [tex]\frac{1}{\lambda}[/tex] as the "wavenumber", whereas in physics, the fraction [tex]\frac{2\pi}{\lambda}[/tex] 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
 
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  • #2
maverick280857 said:
Hi

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

[tex]
\frac{1}{\lambda} = RhcZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})
[/tex]

for the wavelength of radiation emitted when an electron goes from a higher energy level [tex]n_{2}[/tex] to a lower energy level [tex]n_{1}[/tex], 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 [tex]\frac{1}{\lambda}[/tex] as the "wavenumber", whereas in physics, the fraction [tex]\frac{2\pi}{\lambda}[/tex] 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 [tex]\hbar = h/{2\pi}[/tex]. So when physicists use Rydberg's constant, they have to divide it by [tex]2\pi[/tex].
 
  • #3
Also, physicists use the angular frequency [tex] \omega = 2 \pi f[/tex] more often than the regular frequency, f.

So [tex]~\omega = 2\pi f = 2 \pi \frac {c} {\lambda} = c \frac {2\pi} {\lambda}[/tex]

Hence the popularity of that form among physicists.
 
  • #4
Thanks selfAdjoint and Gokul43201

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

Cheers
Vivek
 
  • #5
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,

[tex]\frac{1}{\lambda} = RZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}) \qquad (n_{2} > n_{1})
[/tex]

(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 [tex]\frac{2\pi}{\lambda}[/tex].

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 [tex]n_{1} = 2[/tex] and [tex]n_{2} = 3[/tex], 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 [tex]\frac{2\pi}{\lambda}[/tex].

Cheers
Vivek
 
Last edited:
  • #6
You can write down either [tex] 1/ \lambda ~~or~~2\pi / \lambda [/tex] as long as you provide the correct units to avoid confusion - m^-1 for the former and rad/m for the latter.
 

What is a wavenumber in physics and chemistry?

A wavenumber is a unit of measurement used in physics and chemistry to describe the number of waves that occur in a given distance. It is typically measured in units of inverse meters (m^-1) for physics, and inverse centimeters (cm^-1) for chemistry.

What is the difference between the wavenumber in physics and in chemistry?

While the concept of wavenumber is the same in both physics and chemistry, the units of measurement differ. In physics, the wavenumber is typically measured in inverse meters (m^-1), while in chemistry it is measured in inverse centimeters (cm^-1). Additionally, the wavenumber is often used in different contexts and applications in each field.

Why is the wavenumber important in physics and chemistry?

The wavenumber is a crucial measurement in both physics and chemistry because it allows scientists to quantitatively describe the properties of waves, such as light and sound. It is also used to analyze the behavior and interactions of molecules, atoms, and particles in both fields of study.

How is the wavenumber calculated in physics and chemistry?

The wavenumber is calculated by taking the reciprocal of the wavelength of a wave. In physics, the wavelength is typically measured in meters, while in chemistry it is measured in centimeters. The inverse of these measurements gives the wavenumber in either inverse meters (m^-1) for physics or inverse centimeters (cm^-1) for chemistry.

Can the wavenumber be converted between physics and chemistry?

Yes, the wavenumber can be converted between physics and chemistry by using conversion factors to change the units of measurement. For example, to convert from inverse meters (m^-1) to inverse centimeters (cm^-1), one can multiply the wavenumber value by 100. Conversely, to convert from inverse centimeters (cm^-1) to inverse meters (m^-1), one would divide by 100.

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